First order circuits linear circuit analysis

First Order Circuits
7.1; Introduction
• Let’s analysis circuits comprising of (1) resistor plus capacitor and (2)
resistor plus inductor; called RC and RL circuits, respectively.
• Analysis of RC and RL circuits is done by applying Kirchhoff's laws,
as we did for resistive circuits.
• The only difference is that applying Kirchhoff's laws to purely
resistive circuits results in algebraic equations, while applying these
laws to RC and RL circuits produces differential equations.
• These resulting differential equations are of the first order.

• Hence, these circuits are collectively known as first-order circuits.
• 1st order circuit is characterized by a 1st order differential equation.
• There are two ways to excite these 1st order circuits.
1. Source free circuits contain NO independent sources and relies for
excitation on initial conditions of energy storage elements in circuit.
• We assume that energy initially stored in the ESLs causes current to
flow in the circuit, which is gradually dissipated in the resistors.
• Although source-free circuits are by definition free of independent
sources, they may have dependent sources.

2. Second way of exciting 1st order circuits is by independent sources.
• To begin with, the independent sources we consider are DC sources.
• Later in course, we shall consider sinusoidal & exponential sources.
• The two types of 1st order circuits and the two ways of exciting them
add up to the four possible situations that we will study in this course.
Source-Free RC Circuit
• A source-free RC circuit results on sudden removal of its DC source.
• A source free RC circuit comprising of series combination of a
resistor and an initially charged capacitor, is shown in Fig. 7.1.

• Resistor R and capacitor C in the shown source
free circuit may be the equivalent resistance
and equivalent capacitance respectively.
• We aim to determine the circuit response, which, for instructional
reasons is assumed to be the voltage v(t) across the capacitor.
• Since capacitor is initially charged, we assume that at time t = 0, the
initial voltage is v(t) = VO 7.1
• The corresponding value of the energy stored by the capacitor will be,
w(0) = ½ C VO
2 7.2

• Applying KCL at top node of circuit of Fig. 7.1 yields,
• iC + iR = 0 7.3
• By definition iC = C dv/dt and iR = v/R, substituting in Eq. 7.3, yield,
• C dv/dt + v/R = 0 7.4a.
• Dividing both sides of the Eq. 7.4a by “C” reveal,
• dv/dt + v/RC = 0 7.4b.
• This equation (Eq. 7.4b) is a first-order differential equation, since
only the first derivative of v is involved.
• To solve it, rearrange the terms as dv/v = – (1/RC).dt 7.5.

• Integrating both sides, reveal, ln v = – (t/RC) + ln A,
• Where ln A is the integration constant. Rearranging,
• ln v/A = – t / RC 7.6.
• Taking powers of e produces v(t) = Ae – t/RC.
• From initial conditions {v(t) = A = VO} leaves,
• v(t) = VO e – t/RC. 7.7.
• Equation 7.7 shows that the voltage response of the RC circuit is an
exponential decay of the initial voltage.
• This is referred to as the natural response of the 1st order circuit.

• Natural response is due to (1) initial energy stored and (2) physical
characteristics of the circuit and NOT due to some external source.
• Natural response of a circuit reflects the behavior (in terms of
voltages and currents) of the circuit itself, with no excitation sources.
• The natural response is illustrated graphically in Fig. 7.2.
• Initial condition given by Eq.7.1 i.e.
v(t) = VO, placing t = 0, defines the
graph’s start point.
• Note; as t increases, VO decreases.

• The rapidity with which the voltage decreases is expressed in terms of
the time constant, denoted by  (tau).
• Time constant of a circuit is the time required for the response to
decay to e–1 or 36.8 % of its initial value.
• This implies that at time t = , response v(t) = VO e – t/RC (Eq. 7.7)
becomes VO e-1 = 0.368VO thus  = RC (7.8)
• Hence in terms of the time constant, v(t) = VO e – t/RC (Eq. 7.7) can be
written as v(t) = VO e-t/ (7.9)
• Ratio v(t)/Vo = e-t/ decreases exponentially as reflected in table 7.1.

• It is evident from Table 7.1 that the voltage v(t) is less than 1 percent
of Vo after 5 (five time constants).
• Thus, a capacitor is fully discharged
(or charged) after five time constants.
• In other words, circuit takes 5 to
reach its final or steady state.
• Notice that for all values of t that equals some integer multiple of time
constant , voltage v(t) is reduced by 36.8 % of its previous value i.e.
• v(t + ) = v(t)/e = 0.368v(t) regardless the value of the t.

• From v(t) = VO e-t/ (Eq. 7.9) observe that smaller a time constant,
more rapidly the voltage decreases, that is, faster is the response.
• This is illustrated in Fig. 7.4, reproduced below.
• A small time ( ) constant results in fast response and circuit reaches
the steady state quickly due to quick dissipation of energy stored.

• Conversely a circuit with a large time constant gives a slow response
and takes longer to reach steady (or final) state .
• At any rate, whether the time constant is small or large, the circuit
reaches steady state in five time constants.
• With voltage v(t) = VO e-t/ defined (Eq. 7.9), the current flowing in
resistor can be computed employing Ohm’s law i.e.
• iR(t) = v(t)/R = (VO/R) e–t/ (7.10).
• Power dissipated in resistor, can hence be computed i.e.
• p(t) = viR = (VO
2/R)e–2t/ (7.11).

• Energy absorbed (w = p/t) by the resistor up to time t can be
computed as follows;
•
• Notice that as t  , WR()  ½ CVO
2, which is the same as WC(0),
i.e. the energy initially stored in the capacitor. (refer to equation 7.2)
• Thus the energy that was initially stored in the capacitor is eventually
dissipated in resistor.

• The key to working with a Source-Free RC Circuit is finding:
1) The initial voltage v(O) = VO across the capacitor.
2) The time constant  (sometimes also represented by letter T).
• With above two values, we can determine the circuit response as the
capacitor voltage vC(t) = v(t) = v(0)e-t/.
• Once the capacitor voltage is known, other variables like capacitor
current iC, resistor voltage vR, and current iR can be determined easily.
• Time constant remains same regard less of what the other outputs are.
• How to solve circuits with multiple capacitors and resistors?

• When a circuit contains a single capacitor and several resistors and
dependent sources, then drawing a Thevenin equivalent at the
terminals of the capacitor help form a simple RC circuit.
• Also, one can use Thevenin's theorem when several capacitors can be
combined to form a single equivalent capacitor.
• In finding the time constant  = RC, R is often the Thevenin
equivalent resistance at the terminals of the capacitor; that is, we take
out the capacitor C and find R = RTh, at its terminals.
• Example 7.1: In the circuit drawn at Fig 7.5, let vC(0) = 15 V.

• Find vC, vX and iX at t > 0?
• Solution: We first need to make the circuit
in Fig. 7.5 conform with the standard RC
circuit, recall circuit of Fig. 7.1.
• To do so we find the Thevenin resistance
at the capacitor terminals.
• With capacitor removed, the circuit would
appear as shown in Fig 7.6.
• Compute the Req = (12 + 8) || 5 = 4 .

• Hence the circuit is reduced to one shown in Fig 7.6.
• Computing first the time constant  =RC =4 (0.1)= 0.4 s.
• We can now find; vC = v = v(0)e-t/ = 15e-t/0.4 = 15e-2.5t V.
• Applying voltage division rule (VDR) to circuit of Fig 7.5 yield vx i.e.
• vx = (12/20)(15e-2.5t) = 9e-2.5t V.
• Applying the Ohm’s Law, current; ix = vx /R = 9e-2.5t /12 = 0.75e-2.5t A.
• Problem 7.1: In the circuit at Fig 7.81,
• Voltage; v(t) = 56e-200t V, t > 0.
• And current; i(t) = 8e-200t mA, t > 0.

• Find the values of R & C, time constant & time required for the
voltage to decay half its initial value at t = 0.
• From given values of v(t) and i(t), it is obvious that  = 1/200.
• Resistance R = v(t) / i(t) = 56e-200t V / 8e-200t mA = 56 / 8 = 7 k.
• From  = RC, we find C = /R = 1/(200 x 7000) = 0.7143 F.
• Since initial voltage value at t = 0 is 56, we can compute time of its
decay to half value by equating and solving 1/2 x 56 = 56e-200t
.
• 1/2 x 56 = 56e-200t  e200t = 2  200t0= ln2.
• Hence t0 = ln2 / 200 = 3.466 ms.

• Problem 7.2: Find time constant for the RC circuit of Fig. 7.82.
• As RTh = (120 || 80) +12 = 60 .
• Thus  = 60 x 200 x 10-3 = 12 s.
• Problem 7.3: Find time constant for the RC circuit of Fig. 7.83.
• RTh = 10 + {40 || (20 +30)} = 32.22 k.
•  = RC = (32.22 x 10-3) x (100 x 10-12) = 3.22 s.

• P_Problem 7.1: In Fig. 7.7 let vC(0) = 60 V.
• Find vC, vX, and iO for t  0?
• RTH = (12 || 6) + 8 = 12 .
• Time constant  =RC =12 (1/3)= 4 s.
• So, vC = v = v(0)e-t/ = 60e-t/4 = 60e-0.25t V.
• As 12||6 = 4, applying voltage division,
• vx ={(4)/(4+8)}(60e-0.25t) = 20e-0.25t V.
• Applying KVL to mesh through capacitor,
• v8 = vc – vx = 60e-0.25t – 20e-0.25t = 40e-0.25t V.
• iO = – v8 /R = – 40e-0.25t /8 = – 5e-0.25t A.

• Problem 7.4: Switch in Fig. 7.84 is in position A for a long time.
• If switch moves instantaneously from A to B at t = 0, find v for t > 0.
• For t < 0, v(0 –) = 40 V in single loop.
• For t > 0, it is source-free RC circuit.
•  = RC = 2 x103 x10 x10-6 = 0.02 s.
• Note v(0) = v(0 –) = 40 V. Why?
• v(t) = v(0)e-t/ = 40e-50t V.
• Example 7.2: Switch in circuit of
Fig. 7.8 remained closed and it is
opened at t = 0.

• Find v(t) for t  0 and WC(0) at t = 0.
• Solution: For t < 0, the switch is closed;
the charged capacitor is an open circuit to
DC, as represented in Fig. 7.9(a).
• Using VDR, vC(t) = (9/12)(20)=15V,t<0.
• Voltage across capacitor cannot change instantaneously, hence voltage
across the capacitor at t = 0– is the same at t = 0.
• Therefore, vC(0) = V0 = 15V.
• At, t > 0, when switch is opened, circuit is source free, see Fig. 7.9(b).

• Hence with capacitor removed, RTH = 9 + 1 = 10 .
• Time constant  =RC =10 (20 x 10-3)= 0.2 s.
• Therefore, at t > 0, v(t) = vC(0)e-t/ = 15e-t/0.2 = 15e-5t V.
• Initial energy stored, WC(0) =1/2 Cv2 = 0.5 (20mF)(15)2 = 2.25 j.
• P_Problem 7.2: Switch in circuit of Fig. 7.10 opens at t = 0.
• Find v(t) for t > 0 and wC(0).
• Solution: For t < 0, the switch is
closed; charged capacitor is open
circuit to DC.

• Using VDR, v(12||4) = vC(t) = {(12||4=3)/9)(24)} = 8 V, t < 0.
• As capacitor voltage cannot change instantaneously, hence voltage
across the capacitor at t = 0 – is the same at t = 0 i.e. vC(0) = V0 = 8 V.
• At, t > 0, when switch is opened, circuit behaves as source free circuit.
• Hence with capacitor removed, RTH = 12 || 4 = 3 .
• Time constant  =RC =3 (1/6)= 0.5 s.
• Therefore, at t > 0, v(t) = vC(0)e-t/ = 8e-t/0.5 = 8e-2t V.
• Initial energy stored, WC(0) = 0.5 (1/6)(8)2 = 5.333 j.
• Problem 7.8: In Fig. 7.88, let v = 10e-4t V and i = 0.2e-4t A, at t > 0.

1. Find R, C and , the time constant.
2. Calculate initial energy in the capacitor.
3. Obtain the time it takes to dissipate 50
percent of the initial energy.
• Time constant  =RC = ¼ s. How?
• (from given value of v = 10e-4t V and i = 0.2e-4t A).
• Current – i = C(dv/dt), putting the known values.
• – 0.2e-4t = C(10)(– 4)e-4t  C = 5mF.
• R = /C = 0.25 / 0.005 = 50  and also R = 10e-4t / 0.2e-4t = 50 .

• Class Work; What is the alternative method of finding C & R values.
• Answer: R = v/i = 10e-4t / 0.2e-4t = 50  & C = /R = 0.25/50 = 5 mF.
• The initial energy stored by the capacitor is given by,
• WC(0) = ½ CVO
2 = 0.5 (5 x 10–3)(10)2 = 250 mJ.
• Time it takes to dissipate 50 percent of the initial energy is given by,
• WR = ½ x ½ CVO
2 = ½ CVO
2(1 – e-2to/), solving it yield,
• 0.5 = 1 – e-8to  e-8to = ½  e8to = 2
• And finally tO = 1/8 ln(2) = 86.6 ms.
• Home Work; Solve problems 7.5. 7.6, 7.9 and 7.10 on page-302.

Source-Free RL Circuit
• Source free RL circuit is a resistor & inductor,
connected in series. as shown in Fig. 7.11.
• Since, inductor current cannot change instantly
• Hence we find current i(t) through the inductor as the circuit response.
• At t = 0, i(0) = I0, where I0 is initial current, (7.13).
• Also at t = 0, the energy stored w(0) = 1/2LI0
2 (7.14).
• Applying KVL around the loop in Fig. 7.11, vL+ vR = 0 (7.15).
• As vL = L di/dt and vR = iR, substituting these values in eq. 7.15 yield,

• L di/dt +Ri = 0 or di/dt +(R/L)i = 0 (7.16).
• Rearranging & integrating reveal, ln{i(t)/IO} = – Rt/L (7.17).
• Taking the powers of e, leaves behind, i(t) = IOe-Rt/L (7.18).
• This shows that the natural response of the RL circuit is an exponen-
tial decay of the initial current.
• Current response is shown in Fig. 7.12.
• It is evident from Eq. (7.18) that the
time constant for the RL circuit is,
• Time Constant,  = L/R (7.19).

• Eq. (7.18) turns i(t) = IOe-t/ (7.20).
• From current in Eq. (7.20), we can find voltage across resistor, i.e.
• vR(t) = iR = IORe-t/ (7.21).
• Then power dissipated in the resistor is,
• p = vRi = IORe-2t/ (7.22).
• The energy absorbed by the resistor is.
•

• Note that as t   , wR() = 1/2 L IO
2, which is the same as wL(0), the
initial energy stored in the inductor, see Eq. (7.14).
• So as in case of RC circuits, in RL circuits too, the energy initially
stored in ESL (inductor) is eventually dissipated in the resistor.
• The key to working with a Source-Free RL Circuit is to find:
1) The initial current i(0) = IO through the inductor.
2) The time constant  of the circuit.
• Above two values, help compute the circuit response in terms of iL(t).
• The inductor current iL(t) = i(t) = i(0)e-t/.

• Once the inductor current iL is known, other variables, like inductor
voltage vL, resistor voltage vR, and resistor current iR can be obtained.
• Note that, in general, R in Eq. (7.19) i.e.  = L/R, is the Thevenin
resistance at the terminals of the inductor.
• When a circuit has a single inductor and several resistors and
dependent sources, the Thevenin equivalent can be found at the
terminals of the inductor to form a simple RL circuit.
• Also, one can use Thevenin's theorem to combine several inductors to
form a single equivalent inductor.

• Example 7.3: Let i(0) = 10 A in circuit of
Fig. 7.13, compute currents iX(t) and i(t)?
• There are two ways to solve this problem.
• One way is to obtain the equivalent resistance at the inductor
terminals and then use Eq. (7.20) i.e. i(t) = IOe-t/ .
• Other way is to start from scratch by using Kirchhoff's Voltage Law.
• In either approach aim remains to first compute the inductor current.
• METHOD 1; Find the equivalent resistance at the inductor terminals,
which is the same as the Thevenin resistance.

• Because of the dependent source
presence, insert a voltage source,
v0 = 1 V, at the inductor terminals
a-b, as shown in Fig. 7.14(a).
• Note that a 1-A current source could have been also inserted instead.
• Applying KVL to the two drawn meshes results KVL equations i.e,
• For mesh-1; 2 (i1 – i2) + 1 = 0  i1 – i2 = – 1/2 (7.3.1).
• For mesh-2; 6i2 – 2i1 – 3i1 = 0  i2 = 5/6 i1 (7.3.2).
• Substituting Eq. (7.3.2) into Eq. (7.3.1) yields i1 = – 3 A

• Therefore, current io= – i1= 3A.
• By Ohm’s law, Req = RTh = v0 / i0 = 1/3 ..
• The time constant  is,  = L/R = (1/2) / (1/3) = 3/2 s.
• Thus, the current through inductor is i(t) = i(0)e-t/ = 10e-(2/3)t A, t > 0.
• METHOD 2: Directly apply mesh analysis to circuit of Fig. 7.14(b).
• For mesh-1, {1/2(di1/dt)}+2(il – i2) = 0.
• Multiplying both sides by 2 yields,
• (di1/dt) + 4il – 4i2 = 0 (7.3.3).
• In mesh-2, 6i2 – 2i1 – 3i1 = 0

• Which can also be written as; i2 = 5/6 i1 (7.3.4).
• Substituting Eq. (7.3.4) into Eq. (7.3.3); (di1/dt) + 4i - 4(5/6)il = 0 
• (di1/dt) + (2/3)il = 0  (di1/i1) = – (2/3)dt
• But as i1 = i, therefore, it can be re-written as; (di/i) = – (2/3)dt .
• Next integrating it, and taking powers of e, reveal i(t)= i(0)e–(2/3)t.
• As i(0) = 10 A, therefore, current through the inductor i(t) = 10e–(2/3)t.
• In either case value of current through inductor obtained is same.
• Since inductor voltage, vL = L(di/dt) = 0.5(10)(–2/3)e–(2/3)t = –(10/3)e–(2/3)t.
• Therefore, iX(t) = v/R ={ –(10/3)e–(2/3)t} / 2 = – 1.6667e–(2/3)t, t > 0.

• Home Work: P_Problem 7.3.
• Example 7.4: In circuit of Fig 7.16,
calculate i(t) for t > 0?
• When t < 0, switch is closed, inductor
acts as a short circuit to DC.
• This short circuit 16  resistor & circuit
is reduced to one shown in Fig. 7.17(a).
• As RT = 2 + (12 || 4) = 5 , hence
current iT = i1 = V/RT = 40/5 = 8 A.

• Applying the current division rule (CDR) reveal,
• i(t) = 8 x (12/16) = 6 A, t < 0.
• Since the current through an inductor cannot change instantaneously,
hence i(0) = i(0 –) = 6 A.
• When t > 0, the switch is open and the voltage source is disconnected.
• Revealing a source-free RL circuit shown in Fig. 7.17(b).
• Combining the resistors, reveal; Req= (12+4) || 16 = 8 .
• We can now find the time constant as,  = L/R = 2/8 = 1/4 s.
• Therefore, i(t) = i(0)e – t/ = 6e – 4t A.

• Example 7.5: In circuit of Fig. 7.19, find iO,
vO, and i for all time t i.e. 0  t > 0?
• Assume switch is open for a long time.
• As before, we compute inductor current first
and then other values from it.
• For t < 0, switch is open, inductor conducts &
short circuit the 6  resistor.
• The circuit is reduced to one in Fig. 7.20(a).
• Thus current in 6  resistor; iO = 0, t < 0.

• Applying the Ohm’s law, i = iT = V/RT = 10/(2 + 3) = 2 A, t < 0.
• We can now find voltage vo = v3 = i(t)R = 2 x 3 = 6 V, t < 0.
• At t = 0, inductor current can’t change instantly, hence initial value of
current in the circuit is i(0) = 2A.
• At t > 0, closed switch yield source-free RL circuit of Fig. 7.20(b).
• At the inductor terminals, RTh,= 3 || 6 = 2 .
• And the time constant,  = L/RTh = 2/2 = 1 s.
• Therefore, i = i(t) = i(0)e – t/ = 2e – t/1 = 2e – t A, t > 0.
• Since inductor, 6  and 3  resistors are in parallel, see Fig. 7.20(b),

• Thus vO = vo(t) = – vL = – L (di/dt) = – 2(– 2e– t) = 4e– t V, t > 0.
• And iO = io(t) = vL/R = 4e– t / 6 = – 2/3 e – t A, t > 0.
• Thus for all time,
•
•
• And,
• Fig 7.21 plots i and iO.
• Notice that at t = 0, the inductor current (i) is continuous.

• While at t = 0 current in 6  resistor (iO) drops from 0 to – 2/3 A and
the voltage across the 3  resistor (vO) drops from 6 to 4 V.
• Note that irrespective of the outputs, the time constant remains same.
• Problem 7.11: For the circuit
in Fig. 7.91, find iO for t > 0.
• For t < 0, applying source
transformation circuit can be
rearranged as shown on RHS.

• Inductor less branch Req = 4||4 = 2.
• Converting into two branch circuit, and,
• Applying current division rule, iO(0–) = [2/(2+8)]6 = 1.2 A.
• For t > 0, its source-free RL circuit and  = L/R = 4/12 = 1/3 s.
• Inductor current can’t change instantly, so iO(0–) = iO(0) = 1.2 A.
• Thus, iO(t) = IOe-t/ = 1.2e–3t A, for t > 0.
• Problem 7.13: In circuit Fig. 7.93, let
v(t) = 80e-103t V and i(t) 5e-103t mA, for
t > 0. Find L, R and ?

• Time constant  = 1/103 = 1 ms
• Voltage across the resistor (& parallel inductor) is v(t) = i(t)R.
• Putting the values, 80e–1000t V = R5e–1000t x10–3.
• Sloving it yields, R = (80e–1000t)/(5e–1000t x 10-3)= 16 kΩ.
• From τ = L/R, we find, L = R = (10–3)(16 x 103) = 16 H.
• Home Work: solve problems 7.14,
7.15 and 7.16 on page-303.
• Problem 7.17: In circuit of Fig. 7.97,
find vo(t) if i(0) = 6 A & v(t) = 0.

• RTh = 3+1 = 4  and hence time constant  = L/R = 0.25/4 = 1/16 s.
• When i(0) = 6 A, then i(t) = IOe-t/ = 6e-16t .
• Applying KVL vO(t) = 3i + L(di/dt)
• Putting the values and solving it, yield,
• vO(t) = (3 x 6e-16t) + {(1/4)(– 16)(6e-16t)}.
• Thus vO(t) = – 6e–16t u(t) V.
• Problem 7.18: In circuit Fig. 7.98,
find vo(t) if i(0) = 5 A & v(t) = 0.

• When v(t) = 0 , the circuit can be
redrawn as shown on right.
• Note that Req = 2 || 3 = 6/5 .
•  = L/R = 0.4/1.2 = 1/3.
• i(t) = i(0)e–t/ = 5e–3t A.
• v(0) = –L(di/dt) = – (0.4)(–3) 5e–3t = 5e–3t V.
• Problem 7.19: In circuit of
Fig. 7.99, find i(t) for t > 0 if
i(0) = 6 A.

• To find RTh replace inductor by a 1 V source as shown above.
• Note that a current source will work equally well.
• Performing super mesh analysis, it is obvious that current i = i1.
• Applying KVL to super mesh reveal, 10i – 1 + 40i2 = 0.
• By constraint, i = i2 + i/2,
• Multiplying both sides by 2, reduces it to 2i = 2i2 + i.

• Leading to i = 2i2. or i2 = i/2.
• Putting this value of i2 in super mesh equation 10i – 1 + 40i2 = 0
reveal; 10i – 1 + 20i = 0.
• Therefore, i = 1/30 A.
• We can now find the Rth = v/i = 1  1/30 = 30 .
• Time constant  = L/RTh = 6/30 = 0.2 s.
• And finally current i(t) = IOe-t/ = 6e-5t u(t) A.
• Home Work: Solve problems 7.20, 7.21, 7,22 and 7.23 on page 304.
• Fresh editing required to add double space in all slides ahead of this.

Singularity (Switching) Function:
• Singularity functions (also called switching functions) are functions
that are either discontinuous or have discontinuous derivatives.
• Singularity functions are used commonly in circuit analysis as these
provide good approximations to the switching signals produced in
practical circuits during switching operations
• Response of first-order circuits to a sudden application of
independent DC source can best be comprehended with the help of
singularity or switching functions.

• Singularity functions are helpful in the neat & compact description of
some circuit phenomena, especially the step response of RC or RL
circuits to be discussed later in course.
• The three most widely used singularity functions in circuit analysis
are the unit step, the unit impulse, and the unit ramp functions.
Unit Step Function u(t);
• Unit step function u(t) is 0 for negative and 1 for positive values of t.
• In mathematical terms, (7.24).
• At t = 0, u(t) is undefined, where it changes abruptly from 0 to 1.

• Figure 7.23, depicts the unit step function.
• Notice the abrupt change from 0 to 1 at t = 0.
• Instead of t = 0, abrupt change can occur at t = t0
(where t0 > 0), or at t = – t0, (where – t0 < 0), as
shown in Fig. 7.24 (a & b) respectively.
• In both cases the unit step function becomes,
• (7.25).
• (7.25).
• This is the same as saying u (t) is delayed or
advanced by t0 seconds.

• Step function can reflect an abrupt change in voltage or current.
• For example, sudden application of voltage expressible as unit step
function; v(t) = V0u(t – to) 7.28.
• If we let t0 = 0, then v(t) is simply the step voltage V0u(t).
• Voltage source V0u(t) & its equivalent circuit are shown in Fig. 7.25.
• Note that for t < 0 terminals a-b are short-circuited in equivalent
circuit, yielding vab = 0.
• For t > 0, vab = Vo, appears across
terminals a-b connected to source.

• A current source I0u(t) and its equivalent circuit is drawn in Fig. 7.26.
• Notice that in equivalent circuit, for t < 0, there is an open circuit,
between terminals a-b, which satisfies i = 0 at t < 0.
• Derivative of unit step function u(t) is the unit impulse function (t),
also known as delta function. Mathematically,
• 7.29.
• For t > 0, as expected, i = i0
flows between terminals a-b.
Unit Impulse Function (t):

• Unit impulse function (t) is shown in Fig. 7.27.
• Unit impulse function (t) is zero everywhere
except at t = 0, where it is undefined.
• Impulsive currents and voltages occur in electric circuits as a result of
switching operations or impulsive sources.
• Although the unit impulse function is not physically realizable (just
like ideal sources, resistors, etc.), it is a very useful mathematical tool.
• The unit impulse may be regarded as an applied or resulting shock.
• It may be visualized as a very short duration pulse of unit area.

• For this reason, it is customary to write 1 (denoting unit area) beside
the arrow that is used to symbolize the unit impulse function.
• This may be expressed mathematically as,
• (7.30).
• Note that, t = 0 – denotes the time just before t = 0 and t = 0+ is the
time just after t = 0.
• The unit area is known as the strength of the impulse function.
• When an impulse function has a strength other than unity, the area of
the impulse is equal to its strength.

• For example, Impulse function
10(t) in Fig. 7.28 has an area of 10.
• What is the area of impulse
functions 5(t + 2) and – 4 (t – 3)?
• To illustrate how the impulse function affects integrated functions, let
us evaluate the integral, (7.31).
• Where a < to < b.
• Since (t - to) = 0, except at t = to, hence integrand is “0” except at to.
• Thus, it can be stated that, (7.32).

• Equation 7.32 shows, that when a function is integrated with the
impulse function, we obtain the value of the function only at the point
where the impulse occurs.
• This is a highly useful property of the impulse function known as the
sampling or sifting property.
Unit Ramp Function r(t):
• Integrating unit step function (u(t)} result in unit ramp function r(t) i.e.
• (7.34).
• Mathematically; (7.35).

• The unit ramp function is zero for negative values of time and has a
unit slope t for positive values of time.
• Figure 7.29 shows the unit ramp function.
• Note that a ramp function changes at a
constant rate.
• The unit ramp function may be delayed or advanced & expressed as;
• (7.36).
• (7.37).
• Unit ramp function delayed & advanced by to are shown in Fig 7.30.

• Although there are many more
singularity functions, we are only inter-
ested in the three (the impulse function,
the unit step function, and the ramp
function) at this point.
• Example 7.6: Express the voltage pulse
of Fig. 7.31 in terms of unit step u(t).
• Also calculate its derivative & sketch it.
• Shown pulse represents a gate function.

• It may be regarded as a step function that
switches ON at one value of t and switches
OFF at another value of t.
• Gate function shown in Fig. 7.31 switches ON at t = 2 s and switches
OFF at t = 5 s.
• Its essentially is sum of two step functions as shown in Fig. 7.32(a).

• From the Fig 7.32(a), it is evident that in terms of unit step function,
• v(t) = 10u(t – 2) – 10u(t – 5) = 10[u(t – 2) –u(t – 5)].
• Let’s take its derivative to convert it into impulse function & sketch it.
• Taking the derivative of this yield, = 10[(t – 2) –(t – 5)] .
• Graphically the resulting impulse function is shown in Fig. 7.32(b).
• We can also draw it directly from Fig. 7.31.
• Observing the sudden increase by 10 V at t = 2 s
leading to 10(t - 2) & sudden decrease by 10 V
at t = 5 s yielding -10(t - 5).

• Class Work: Draw wave form for
singularity functions,
• i(t) = [u(t + 2) + u(t – 2)] A, and,
• v(t) = [r(t) – r(t–3) + 4u(t–5) – 7u(t–8)] V.
• Example 7.7: Express the saw tooth
function shown in Fig. 7.35 in terms of
singularity functions.
• It can be solved in more than one way.

• Easiest approach is based on mere observation of the given function.
• Look at sketch of v(t) in Fig. 7.35; it is not hard to notice that v(t) is
essentially a combination of multiple singularity functions.
• So let v(t) = vl(t) + v2(t) + …. (7.7.1).
• Function v1(t) is ramp function of slope 5, as shown in Fig. 7.36(a).

• Hence, v1(t) = 5r(t) (7.7.2).
• As vl(t) goes to infinity, we need another function at t = 2 s to stop it
from rising indefinitely.
• Let this function be v2(t); a ramp function of slope (–5), as shown in
Fig. 7.36(b).
• Thus, v2(t) = -5r(t - 2) (7.7.3).
• Adding vl and v2 gives us the signal shown in Fig. 7.36(c).
• Obviously, this is not the same as v(t) shown in Fig. 7.35.
• To conform to v(t), the signal must drop by constant 10 units at t = 2 s.

• Thus adding a third signal v3, where, v3 = -10u(t - 2) (7.7.4),
we get v(t), as shown in Fig. 7.37.
• Substituting Eqs. (7.7.2) through (7.7.4) into Eq. (7.7.1) finally gives,
• v(t) = vl(t) + v2(t) + v3(t) = 5r(t) - 5r(t - 2) - 10u(t - 2).

• Method-2: A close observation of Fig. 7.35 reveals that v(t) is a
multiplication of two functions: a ramp function and a gate function.
• Thus, v(t) = 5t[u(t) - u(t - 2)]
• v(t) = 5tu(t) - 5tu(t - 2), once rearranged,
• v(t) = 5r(t) - 5(t - 2 + 2)u(t - 2)
• v(t) = 5r(t) - 5(t - 2)u(t - 2) - 10u(t - 2)
• v(t) = 5r(t) - 5r(t - 2) - 10u(t - 2), this result is the same as before.
• Self Study: method-3, seeing v(t) as product of ramp & step functions.
• Class Work: P_Problem 7.7.

• Example 7.8: For the given signal express
g(t) in terms of step and ramp functions?
• The signal g(t) may be regarded as the sum of three functions specified
within the three intervals t < 0, 0 < t < 1, and t > 1.
• For t < 0, signal g(t) may be regarded as 3 multiplied by u(-t), where
u(-t) = 1 for t < 0 and 0 for t > 0.
• Within the time interval 0 < t < 1, function may be considered as -2
multiplied by a gated function [u(t) - u(t - 1)].

• For t > 1, the function may be regarded as 2t - 4 multiplied by the unit
step function u(t - 1).
• Resulting linear combination of singularity function for the given
signal is, g(t) = 3u(– t) – 2[u(t) – u(t – 1)] + (2t – 4)u(t – 1),
• Once rearranged, g(t) = 3u(– t) – 2u(t) + 2u(t – 1) + (2t – 4)u(t – 1),
• Combining last 2 terms, g(t) = 3u(– t) – 2u(t) + (2t – 4 + 2)u(t – 1),
• Solving it reveal, g(t) = 3u(– t) – 2u(t) + (2t – 2 )u(t – 1),
• Again rearranged into, g(t) = 3u(– t) – 2u(t) + 2(t – 1)u(t – 1),
• As t.u = r, ramp function, a singular function representation becomes,

• g(t) = 3u(– t) – 2u(t) + 2r(t – 1),
• Which is the singular function representation in terms of step and
ramp functions.
• It is also possible to avoid use of u (– t) by replacing it with 1 – u(t).
• Then, g(t) = 3[1 – u (t)] – 2u (t) + 2r (t – 1), which on expanding is,
• g(t) = 3 – 3u (t) – 2u (t) + 2r (t – 1), which once simplified yields,
• g(t) = 3 – 5u (t) + 2r (t – 1), another better compact representation.
• Home Work; plot g(t) and apply Method 1 used in Example 7.7, to
write the linear combination of corresponding singularity functions.

• Problem 7.24: Express below signals in terms of singularity functions.
a. Signal
• v(t) = – 5u(t)
b. Signal
• i(t) = – 10[u(t – 1) – u(t – 3)] + 10[u(t – 3) – u(t – 5)], or, once
simplified, i(t) = – 10u(t – 1) + 20u(t – 3) – 10[u(t – 5).
c. Signal
• x(t) = (t – 1) [u(t – 1) – u(t – 2)] + 1[u(t – 2) – u(t – 3)] + (4 – t) [u(t – 3) – u(t – 4)].

• x(t) = (t – 1)u(t – 1) – (t – 1)u(t – 2) + u(t – 2) – u(t – 3) + (4 – t)u(t – 3) – (4 – t)u(t – 4).
• x(t) = (t – 1)u(t – 1) –u(t – 2) (t – 1 – 1) – u(t – 3)(+1 – 4 + t) – (4 – t)u(t – 4).
• x(t) = (t – 1)u(t – 1) – (t – 2)u(t – 2) – (t – 3)u(t – 3) – (t – 4)u(t – 4).
• x(t) = r(t – 1) – r(t – 2) – r (t – 3) + r (t – 4).
d. Signal
• y(t) = 2u(– t) – 5[u(t) – u (t – 1)].
• Singularity functions above can also be expressed as,
• y(t) = 2u(– t) – 5u(t) – 5u (t – 1).

• Problem 7.26: Express below signals in terms of singularity functions.
a. v1(t) = u(t + 1) – 2u(t) + u(t – 1).
b. v2(t) = 2u(t – 2) – r(t – 2) + r(t – 4).
c. v3(t) = 2u(t – 2) + 2u(t – 4) – 4u(t – 6).

• Problem 7.27: Express v(t) in Fig. 7.105 in terms of step functions.
• v(t) = [5u(t+1) + 10u(t) – 25u(t –1) + 15u(t – 2)] V.
• Problem 7.28: Sketch a waveform represented by singularity function;
i(t) = r(t) – r(t – 1) – u (t – 2) – r (t – 2) + r (t – 3) + u (t – 4)?
• Once sketched i(t) would appear as shown below right corner.

Step Response of an RC Circuit
• Sudden application of DC source to an RC circuit, may that be
voltage or current source, can be modeled as a step function.
• Behavior of circuit when the excitation is a step
function, which may be voltage or current
source, is called step response.
• Consider the RC circuit of Fig. 7.40(a) where Vs
is a constant DC voltage source, which can be
replaced by the circuit in Fig. 7.40(b).

• As before, we select the capacitor voltage as the circuit response to
be determined and assume an initial voltage (Vo) on the capacitor.
• Since the voltage of a capacitor cannot change instantaneously, thus,
• (7.40).
• Where v(0-) is the voltage across the capacitor just before switching
and v(0+) is its voltage immediately after switching.
• This is same as saying voltage v at t < 0 is same as at t = 0.
• Applying KCL to circuit of fig 7.40b reveal,
• , where v is the voltage across the capacitor.

• Which can be rearranged as, (dividing both sides by C)
• (7.41).
• For t > 0, u(t) = 1, hence, Eq. (7.41) becomes,
• (7.42).
• Rearranging the terms leads to equation,
•
• Which can further be rearranged into,
• (7.43).

• Integrating both sides and introducing initial conditions,
• or
• Alternatively, (7.44).
• Taking the exponential of both sides where  = RC.
• Rearranging it again reveal,
• We can now compute the value v at t > 0 i.e. v(t),
• v(t) = VS + (VO – VS)e–t/, t > 0 (7.45).
• Thus, (7.46).

• This (Eq. 7.45) is known as the complete response (or total response)
of the RC circuit to a sudden application of a DC voltage source,
(assuming that the capacitor is initially charged).
• Assuming that Vs > Vo, a plot of v(t) is shown in Fig. 7.41

• If we assume that the capacitor is uncharged initially, we set VO = 0 in
Eq. (7.46) yielding;
• (7.47).
• Alternatively, it can be written as; v(t) = Vs(1 - e-t/)u(t) (7.48).
• This (Eq. 7.48) is the complete step response of the RC circuit when
the capacitor is initially uncharged.
• Using Eq. 7.47 and Ohm’s Law, current through capacitor is given by,
• i(t) = C(dv/dt) = (C/ )(Vse-t/),  = RC, t > 0, or,
• i(t) = (Vs/R)e-t/ u(t) (7.49).

• Fig. 7.42, shows plots of capacitor
voltage v(t) and current i(t).
• Rather than going through the above
derivations , there is a short-cut
method-for finding the step response of
an RC (or even RL) circuit.
• Let us re-examine Eq. (7.45), which
states that, v(t) = VS + (VS – VO)e-t/.
• Note that, v(t) has two components.

• There are two ways of decomposing v(t) into two components.
• The first is to break it into a "natural response and a forced response“.
• Second is to break it in "transient response & steady-state response."
• To apply option-1, let “v” be a complete response “vn” be natural
response and “vf” be the forced response.
• Splitting complete response into natural and forced response yield
equation v = vn + vf (7.50).
• Where, vn = VOe–t/ is natural response (7.50a).
• And, vf = VS(1 – e–t/) is the forced response (7.50b).

• We are familiar with the natural response vn of the circuit.
• Why vf is known as forced response because it is produced by circuit
when an external "force" (a voltage source in this case) is applied.
• It represents what the circuit is forced to do by the input excitation.
• The natural response eventually dies out along with the transient
component of the forced response, leaving only the steady-state
component of the forced response.
• Another way of looking at the complete response is to break into two
components; one temporary and the other permanent.

• If “v” again represent a complete response then let its two components
be “vt” the temporary part and “vSS” be the permanent part.
• Splitting complete response into transient and steady state responses
yield equation v = vt + vSS (7.51).
• Where, vt = (VO – VS)e–t/ (7.52a).
• And, vSS = VS (7.52b).
• The transient response vt is temporary; it is the portion of the
complete response that decays to zero as time approaches infinity.

• The steady-state response vss, is the permanent portion of complete
response that remains after the transient response has died out.
• The steady-state response is the behavior of the circuit a long time
after an external excitation is applied.
• The first decomposition of the complete response is in terms of the
source of the responses, while the second decomposition is in terms of
the permanency of the responses.
• Under certain conditions, natural response & transient response are
the same (i.e. also true for forced response & steady-state response).

• Whichever way we look at it, the complete response in Eq. (7.45) may
be written as: v(t) = v() + [v(0) - v()]e-t/ 7.53
• Where, v(0) is the initial voltage at t = 0+ and v() is the final or
steady-state value.
• Thus, to find the step response of an RC circuit requires three things:
1. The initial capacitor voltage v(0).
2. The final capacitor voltage v().
3. The time constant .
• We obtain v(0) for t < 0 plus v() &  for t > 0 from the given circuit.

• Once v(0), v() &  are known, response is computed by Eq. (7.53).
• Note that if the switch changes position at time t = t0, instead of t = 0,
there is a time delay in the response so that Eq. (7.53) becomes,
v(t) = v() + [v(t0) - v()]e-(t-t0
)/ (7.54).
• Where v(t0) is the initial value at t = t0 .
• Keep in mind that Eq. (7.53) or (7.54) applies only to step responses,
that is, when the input excitation is constant.
• Example 7.10: The switch in circuit of Fig. 7.43 has been in position
A for a long time. At t = 0, the switch moves to position B.

• Charged capacitor acts like an open circuit to DC, but v; the voltage
across the capacitor remains same as voltage across the 5-k resistor.
• Hence, the voltage across the capacitor just before t = 0 is obtained by
voltage division rule i.e. v(0–) = (5/8) x 24 = 15 V.
• We can conclude that, v(0) = v(0-) = v(0+) = 15 V. Why?
• Determine v(t) for t > 0, and its
value at t = 1 s plus t = 4 s?
• For t < 0, the switch is at
position A.

• Answer; Since the capacitor voltage cannot change instantaneously.
• For t > 0, the switch is in position B.
• Thevenin resistance connected to the capacitor; RTh = 4 k.
• Therefore, the time constant;  =RThC= (4 x103)(0.5 x 10-3) = 2s.
• As capacitor is open circuit to DC at steady state, thus v() = 30 V.
• Applying the formula, v(t) = v() + [v(0) - v()]e-t/ & putting values,
• v(t) = 30 + (15 - 30)e-t/2 = (30 - 15e-0.5t) V.
• At t = 1 s; v(1) = (30 - 15e-0.5t) = (30 - 15e-0.5) = 20.90 V.
• At t = 4 s; v(4) = (30 - 15e-0.5t) = (30 - 15e-2) = 27.97 V.

• Example 7.11: In Fig. 7.45, the switch has been closed for a long time
and is opened at t = 0. Find i and v for all time?
• Note that the resistor current i can be discontinuous at t = 0, while the
capacitor voltage v cannot.
• Hence, it is always better to find v first and then obtain i from v.

• By definition of the unit step function,
• For t < 0, the switch is closed and
30u(t) = 0, hence it can be replaced by
a short circuit, as in Fig. 7.46(a).
• With switch closed for t < 0, capacitor
voltage v has reached steady state and
capacitor is an open circuit, as shown.
• Note; resistors & source are in parallel.
• Hence in Fig.7.46(a), v10 = 10 V & i = –v/R = –10/10 = –1 A.

• At t = 0, since the capacitor voltage cannot change instantaneously,
therefore it can be concluded that, v(0) = v(0–) =10V.
• For t > 0, the switch is opened and the 10 V voltage source is
disconnected from the circuit.
• The 30u(t) voltage source is now operative, as shown in Fig. 7.46(b).
• After a some time, the circuit reaches steady state and the capacitor
acts like an open circuit again.
• Hence, v() by using voltage division is, v() = (20/30) x 30 = 20 V.
• We next compute RTh = 10 || 20 = 200/30 = 6.667 .

• Hence time constant  = 6.667 x 0.25 = 5/3 s =1.667 s.
• We can now apply Eq. 7.53 i.e. v(t) = v() + [v(0) - v()]e-t/.
• Thus v(t) = = 20 + (10 - 20)e-t/(1.667) = (20 - 10e-0.6t) V.
• To obtain i, we notice from Fig. 7.46(b) that i is the sum of the
currents through the 20- resistor and the capacitor.
• Thus i = (v/20) + C (dv/dt) for t > 0 (a).
• As v = (20 - 10e-0.6t) hence dv/dt = (0 – 10e-0.6t) d/dt ( – 0.6t) or,
• dv/dt = – 10 (– 0.6)e-0.6t = 6e-0.6t, putting these values in Eq. (a),
• Current i = [(20 – 10e-0.6t)/20] + [0.25(6e-0.6t)].

• Once simplified, reveal, i = (1 – 0.5e-o.6t) + 1.5e - 0.6t = (1 + e-0.6) A.
• In fig. 7.46(b) v + 10i = (20 - 10e-0.6t) + 10(1 + e-0.6) = 30, indicating
that KVL, as expected, is satisfied.
• Notice that the capacitor voltage is continuous while the resistor
current is not.
• Hence, to conclude;
• And current;

• Let capacitor voltage v = v1 + v2,
where v1 is due to the 12 V source
and v2 is due to the 2 A source.
• Converting 2 A current source in parallel
with 4  resistor into 8 V voltage source
in series with 4  resistor; v2 = 8 V.
• Thus at t < 0, voltage v = 12 – 8 = 4 V.
• Problem 7.39: Calculate capacitor voltage v for t < 0 and t > 0 for the
circuit of Fig. 7.106(b).

• At t > 0, closed switch isolates current source & circuit reduces to;
• As v() = 12 , v(0) = 4 ,  = RC = 2 x 3 = 6 s.
• Applying, v(t) = v() + [ v(0) – v()] e-t/
• v(t) = 12 + (4 – 12) e-t/6 = 12 – 8e-t/6 V.
• Problem 7.40: find capacitor voltage v for t < 0 & t > 0 for circuit
shown in Fig. 7.107a.
• For t < 0, capacitor voltage, v = 12 V.
• At t > 0, v(t) = v() + [ v(0) – v()] e-t/.
• Since v() = 4 V, v(0) = 12 V and  = 2 x 3 = 6 s.

• Therefore, at t > 0, v(t) = 4 + (12 – 4) e-t /6 = 4 + 8e-t/6 V.
• Problem 7.42b; Switch in the circuit of Fig. 7.109, (switch modified)
remained closed for a long time and is opened at t = 0. Find vo(t)?
• At t < 0, vC = vO(0) = 12(4/6) = 8 V.
• At t = 0, it is source free RC circuit.
• RTh = 4 , so,  = RC = 4 x 3 = 12s.
• Hence v(t) = v(0) e-t/ = 8 e-t/12 V.
• Problem 7.43: In the circuit of
Fig. 7.110, find i(t) for t < & > 0?

• At t < 0, circuit has reached steady state and capacitor is open circuit.
• Applying KCL at node VO, – 2 + VO/40 – 0.5i + i = 0.
• Once simplified, – 2 + VO/40 + 0.5i = 0, but at same time, i = VO/80.
• Thus, – 2 + VO/40 + 0.5(VO/80) = 0 or 2 = VO/40 + VO/160
• Once solved, VO = 64 V, hence, for t < 0, i = VO/80 = 64/80 = 0.8 A.
• At t > 0, it is a source free circuit as shown in Fig. (b) on next slide.
• Transforming 80 V into 2 A
source, equivalent circuit would
appear as shown in Fig. (a).

• To find RTh, replace the capacitor
with a 1 V voltage source as shown
in Fig. (c). Why? (dependent source)
• KCL at node VC yield, iT = – 0.5i + i.
• But i = VC/80 = 1/80, and thus,
• iT = – 0.5(1/80) + 1/80 = 0.5/80 A.
• RTh = VC/itotal.= 80/0.5 = 160 , so  = RThC = 160 x 0.003 = 480 s.
• Note that, VO(0-) = VO(0) = 64 V.
• Therefore, being a source free circuit, v(t) = vC(t) = VOe-t/ = 64 e-t/480.

• Since, iT = iC = C(dvC/dt) = 0.003(–1/480)64e-t/480 = – 0.4 e-t/480 mA.
• From KCL equation iT = – 0.5i + i, it is obvious that, i = iT/0.5.
• Therefore, i(t) = iT/0.5 = (– 0.4/0.5)e-t/480 = – 0.8 e-t/480 u(t) mA, t > 0.
• Problem 7.44: In the circuit of fig.
7.111 calculate i(t) for all t > 0?
• Note; switch moves a to b at t = 0.
• At t < 0, charged capacitor is open circuit & v(0) = (3/9) x 60 = 20 V.
• At t = 0, 60 V source is removed and 24 V source is applied to circuit.
• Thus at t > 0, when steady state arrives, v() = (3/9) x 24 = 8 V.

• Now RTh = 6||3 = 2  and  = RC = 2 x 2 = 4 s hence v(t) can now be found,
• So, v(t) = v() + [ v(0) – v()] e-t/ = 8 + (20 – 8e-t/4) = 8 + 12e-0.25t.
• So for at all t > 0, i(t) = C(dv/dt) = (2) x (12)(– 0.25)e-0.25t. = – 6 e-0.25t.
• Problem 7.45: Find vO in the circuit of
Fig. 7.112, when vS= 30u(t) V.
• Assume that vO(0) = 5 V.
• Equivalent circuit to find RTh would
appear as shown here.
• Thus RTh = 20||40 + 10 = 23.334 k,

• At t > 0, vS = 30 V, capacitor is
charged and act as short circuit, the
equivalent circuit is drawn here.
• Applying voltage division, the steady state voltage;
• v() = v40 = [40/(40+20)]30 = 20 V.
• Time constant;  = RC = 23.334 k x 3 F = 0.07 s.
• Putting known values in, v(t) = v() + [ v(0) - v()] e-t/.
• v(t) = 20 + [ 5 – 20] e-t/0.07.= 20 – 15 e-14.286t u(t) V.
• Home Work; Solve problems 7.46, 7.47 and 7.48.

Step Response of an RL Circuit
• Modelling the voltage source in RL circuit
of Fig. 7.48(a), as step function, the circuit
may be replaced by Fig. 7.48(b).
• Goal is to find the inductor current i as the
step response of RL circuit.
• Contrary to lengthy RC circuit derivations,
let’s use short cut based on Eqs. (7.50-53)
to find RL circuit step response (current i).

• Let the response be the sum of the transient and the steady-state
responses, i.e. i = it + iss 7.55.
• Recall that the transient response is always decaying exponentially.
• The transient response in terms of current, it = Ae-t/, 7.56.
• Here time constant,  = L/R and A is a constant to be determined.
• The steady-state response is the value of the current a long time after
the switch in Fig. 7.48(a) is closed.
• Transient response dies out after five time constants, at that time, the
inductor becomes a short circuit, and the voltage across it is zero.

• Thus the entire source voltage vs appears across resistor R and the
steady-state response is iss = Vs/R 7.57.
• Substituting Eqs. (7.56) and (7.57) into Eq. (7.55) yields current
response, i = it + iss = Ae-t/ + Vs/R 7.58.
• We now determine the constant A from the initial value of i.
• Let IO be the initial current through the inductor, which may come
from a source other than Vs.
• Since the current through the inductor cannot change instantaneously,
therefore, i(0+) = i(0– ) = I0 7.59.

• Thus, at t = 0, Eq. (7.58) becomes I0 = A + Vs/R.
• From this, we obtain A as, A = I0 – Vs/R and substituting it for A in Eq.
(7.58), we get, i(t)= Vs/R + (I0 – Vs/R) e-t/ 7.60.
• This is the complete response of the RL circuit, which may also be re-
written as, i(t) = i( ) + [i(0) – i( )]e-t/ 7.61.
• Where i(0) & i() are initial & final values of
current i. Fig. 7.49 reflects the total response.
• To find the step response of an RL circuit
requires three things: i(0) at t = 0, i() and .

• Note that i(0) at t = 0 is initial inductor current, which is obtained
from the circuit analysis by determining inductor current at t < 0.
• Final inductor current i() and time constant  are also determined
from the given circuit taking into account switch position at t > 0.
• Once three items are known, the response is obtained using Eq. (7.61).
• Again, if the switching takes place at time t = t0 instead of t = 0, then,
Eq. (7.61) becomes, i(t) = i() + [i(to) - i()]e-(t-to)/ 7.62.
• If I0 = 0, then, 7.63a.
• Alternatively 7.63b.

• This is step response of RL circuit with no initial inductor current.
• The voltage across the inductor is obtained from Eq. (7.63).
• Using v = L (di/dt), we get, v(t) = VS (L/R)e–t/, as  = L/R therefore,
• Inductor voltage v(t) = VS (L/R)(R/L)e–t/ = Vs e-t/ u(t) 7.64.
• Figure 7.50 shows the step responses in Eqs. (7.63) and (7.64).S

• Example 7.12: Find i(t) in the circuit of
Fig. 7.51 for t > 0. Assume that the
switch has been closed for a long time.
• At t < 0, the 3 resistor is short-circuited,
also the inductor acts like a short circuit.
• The current through the inductor at t < 0 is, i(0–) = 10/2 = 5 A.
• Inductor current can’t change abruptly, so, i(0) = i(0+) = i(0–) = 5 A.
• When t > 0, the switch is open and 2  and 3  resistors are in series.
• Hence the final value of inductor current, i() = 10/(2 + 3) = 2 A.

• As the, RTh = 2 + 3 = 5  and  = L/Rth = (1/3)/5 = 1/15 s = 0.0667 s.
• So, i(t) = i() + [i(0) – i()]e-t/ = 2 + [5 – 2]e –15t = 2 + 3e –15t A, t > 0.
• Result can be confirmed by applying KVL and see if it is satisfied.
• Class Work: Check to verify that KVL is satisfied in Fig. 7.51..
• Solution: vS = vR + vL = {i(t) x (RTh)} + {L x (di/dt)}.
• Putting the values, 10 = {(2 + 3e –15t) x (5)} + {(1/3) x (di/dt)}.
• Since for i = (2 + 3e –15t), di/dt = 0 + 3e –15t d/dt (–15) = 3(–15)e –15t
• Therefore, 10 = {(10 + 15e –15t)} + {(1/3) x (3(–15)e –15t)} = 10.
• That confirms the result.

• Example 7.13: At t = 0, switch S1 in
Fig. 7.53 is closed, and switch S2 is
closed 4 s later.
• Find i(t) for t > 0 and calculate i for t = 2 s and t = 5 s.
• We need to consider the three time intervals t  0, 0  t  4, and t  4
separately. For t < 0, switches S1 and S2 are open hence current i = 0.
• Since the inductor current cannot change instantly,
• Therefore; i(0-) = i(0) = i(0+) = 0.

• For 0  t  4, Sl is closed so that the 4  and 6  resistors are in
series. (Remember, at this time, S2 is still open.).
• Assuming for now that S1 is closed forever, i() = 40/(4 + 6) = 4 A.
• Thus, RTh = 4 + 6 = 10  and  = L/Rth = 5/10 = 1/2 s = 0.5 s. Hence,
• i(t) = i() + [i(0) - i()]e-t/ = 4 + [0 – 4]e –2t = 4(1 – e –2t) A, 0  t  4.
• For t  4 , S2 is also closed; the 10 V voltage source is now connected.
• This sudden change does not affect the inductor current because the
current cannot change instantaneously in inductor.
• Thus, initial current at t  4 is i(4) = i(4 –) = 4(1 – e – 8) = 3.998  4 A.

• To find i(), in this case, let v be the voltage at node P in Fig. 7.53.
• Using KCL yield nodal equation, [(v – 40)/4] + [(v – 10)/2] + v/6 = 0.
• Solving it yield v = 16.364 V, therefore i() = 16.364/6 = 2.727 A.
• Since, RTh = 4||2 + 6 = 22/3  and  = L/RTh = 5/(22/3) = 15/22 s.
• Thus, i(t) = i() + [i(0) – i()]e – (t – to)/ = 2.727 + [4 – 2.727]e – (t – 4)/ .
• Not that, (t - tO) in the exponential is added because of the time delay.
• Now replacing value of  = 15/22, the value of i(t) at t  4 becomes,
• i(t) = 2.727 + [4 – 2.727]e –1.4667(t – 4) = 2.727 + 1.273e –1.4667(t – 4), t  4
• Putting all together,

• At t = 2, i(t) = 4(1 – e –2t) = 4(1 – e –4) = 3.926 A.
• For t  4, i(t) = 2.727 + 1.273e –1.4667(t – 4) , therefore,
• At t = 5, i(t) = 2.727 + 1.273e –1.4667(5 – 4) = 3.02 A.
• Problem 7.53: Determine inductor
current i(t) for both t < 0 and t > 0 for
both the circuits in Fig. 7.119.
• In Fig. 7.119a, at t < 0, inductor current
i = 25/(3 + 2) = 5 A.
• At t > 0, i(0) = 5 &  = L/R = 4/2 = 2 s.

• Closed switch at t > 0, isolate3s source from the inductor.
• Inductor current in resulting source free circuit is given by,
• i(t) = i(0) e–t/ = 5 e–0.5t u(t) A.
• In Fig. 7.119b, at t < 0, the inductor short circuit 2  & 4  resistors.
• Thus i(t) = 6 A at t < 0, but at t > 0, we get source free RL circuit.
• As  = L/R = 3/2 s, therefore, i(t) = i(0) e–t/ = 6 e–2t/3 u(t) A.
• Problem 7.61: In circuit of Fig, 7.126 source current iS changes from
5 A to 10 A at t = 0 i.e. iS = 5u(-t) + 5u(t).
• Find v & i in the circuit at t > 0?

• Since RTh = 4 , therefore,  = L/R = 0.5/4 = 0.125 s.
• At t  0 the initial value of current i(0) = 5
• The steady state or final value of current i() = 10 A.
• Substituting the known values in, i(t) = i() + [i(0) – i()]e-t/
• i(t) = 10 + [5 – 10]e –8t = (10 – 5e –8t) u(t) A.
• To find the voltage v put known values in,
• v(t) = L(di/dt) = ½ (-5)(-8) e –8t = 20 e –8t u(t) A.
• Self Study; Applications of 1st order circuits and chapter summary.
• Home Work; P_Problem 7.13, problems 7.59, 7.62, 7.63, and 7.64.

Summary:
o The analysis in chapter-7 is applicable to any circuit that can be
reduced to an equivalent circuit comprising a resistor and a single
energy-storage element (inductor or capacitor).
o Such a circuit is first-order because its behavior is described by a first-
order differential equation.
o When analyzing RC circuits, always keep in mind that the capacitor is
an OPEN circuit to steady-state dc conditions.
o While analyzing RL circuits, always keep in mind that the inductor is
a SHORT circuit to steady-state dc conditions.

 The NATURAL response is obtained when no independent source is
present. It has the general form, x(t) = x(0)e-t/ .
 Where x represents current through (or voltage across) a resistor, a
capacitor, or an inductor, and x(0) is the initial value of x.
 Because most practical resistors, capacitors, and inductors always
have losses, natural response is a transient response. (dies out with time)
 The time constant  or T is the time required for a response to decay to
1/e of its initial value.
 For RC circuits,  = RC and for RL circuits,  = L/R.

• The singularity functions include the unit step, the unit ramp function,
and the unit impulse functions.
• The unit step function u(t) is,
• The unit impulse function is,
• The unit ramp function is,
 The steady-state response is the behavior of the circuit after an
independent source has been applied for a long time.
 The transient response is the component of the complete response that
dies out with time.

 The total or complete response consists of the steady-state response
and the transient response.
o The step response is the response of the circuit to a sudden application
of a dc current or voltage.
o Finding the step response of a first-order circuit requires the initial
value x(0+), the final value x(), and the time constant .
o With these three items, we obtain the step response as,
x(t) = x() + [x(0+) – x()]e – t/
o A more general form of this equation is,
x(t) = x() + [x(to+) – x()]e– (t – to)/
o Or we may write it as
Instantaneous value = Final + [Initial – Final]e – (t – to)/.

• Four practical applications of RC and RL circuits are: a delay circuit, a
photoflash unit, a relay circuit, and an automobile ignition circuit.
• PSpice is very useful for obtaining the transient response of a circuit.
Appendix D provides a review of transient analysis using PSpice.
• Op amp circuit with a storage element exhibit 1st-order behavior.
• In practice inductors are hardly ever used in op amp circuits;
therefore, the op amp circuits encountered the most are the RC type.
• Op amp circuits are analyze using nodal analysis.
• Some-times, the Thevenin equivalent circuit is used to reduce the op
amp circuit to one that we can easily handle.
• Home Assignment: Solve review questions 7.1 to 7.10 (P-300 to 301)
and problems 7.11 to 7.17 (P-301 to 303)

First order circuits linear circuit analysis

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