Signals & Systems Chapter Analysis Using TF

Signals & Systems
Chapter 5
Time-domain and
frequency-domain analysis
of LTI systems using TF

INC212 Signals and Systems : 2 / 2554

Overview
 Stability and the impulse response
 Routh-Hurwitz Stability Test
 Analysis of the Step Response
 Fourier analysis of CT systems
 Response of LTI systems to sinusoidal inputs
 Response of LTI systems to periodic inputs
 Response of LTI systems to nonperiodic
(aperiodic) inputs

INC212 Signals and Systems : 2 / 2554 Chapter 5 The analysis of LTI systems using TF

Stability and
the Impulse Response

bM s M + bM −1s M −1 +  + b1s + b0 Numerator
H ( s) =
a N s N + a N −1s N −1 +  + a1s + a0 Denominator

Assumption : N ≥ M
: H (s) dose not have any common poles and zeros


Stability and

H(s) has a real pole, p h(t) contains cept

H(s) has a complex pair poles, σ±jω
h(t) contains ceσtcos(ωt+θ)

H(s) has a repeated poles
h(t) contains ctiept or ctieσtcos(ωt+θ)


Stability and
h(t)  0 as t  ∞ Stability

cept p<0

ceσtcos(ωt+θ) σ<0

ctiept or ctieσtcos(ωt+θ) p < 0 or σ < 0


Stability and
h(t)  0 as t  ∞ Stable

|h(t)| ≤ c for all t Marginally Stable

bounded

|h(t)|  ∞ as t  ∞ Unstable

unbounded


Stability and
OLHP ORHP

Stable Unstable
× × ×

× × ×

× × ×

Marginally Stable


Stability and
 Example 8.1 Series RLC Circuit

1 LC R
H ( s) = 2 b<0 Re( p1 , p2 ) = − <0
2L
s + ( R L) s + 1 LC
R −
R
+ b <0
p1 , p2 = − ± b
2L b≥0 2L
2
 R 
2 b< 
 R  1  2L 
b=  −
 2 L  LC −
R
− b <0 b<
R
2L 2L


Routh-Hurwitz Stability Test
Table 8.1 Routh Array A( s ) = a N s N + a N −1s N −1 +  + a1s + a0 , a N > 0
ai > 0 for i = 0,1,2,  , N − 1
sN aN aN-2 aN-4 …
sN-1 aN-1 aN-3 aN-5 … a N −1a N − 2 − a N a N −3 a a
bN − 2 = = a N − 2 − N N −3
a N −1 a N −1
sN-2 bN-2 bN-4 bN-6 …
a N −1a N − 4 − a N a N −5 a a
bN − 4 = = a N − 4 − N N −5
sN-3 cN-3 cN-5 cN-7 … a N −1 a N −1
…

…

…

…

bN − 2 a N −3 − a N −1bN − 4 a b
c N −3 = = a N −3 − N −1 N − 4
s2 d2 d0 0 … bN − 2 bN − 2
s1 e1 0 0 … c N −5 =
bN − 2 a N −5 − a N −1bN −6 a b
= a N −5 − N −1 N −6
bN − 2 bN − 2
s0 f0 0 0 …


Table 8.1 Routh Array
If all elements > 0 sN aN aN-2 aN-4 …
Stable sN-1 aN-1 aN-3 aN-5 …
sN-2 bN-2 bN-4 bN-6 …
If 1 or more elements = 0
& no sign changes sN-3 cN-3 cN-5 cN-7 …

…

…

…

…
Marginally Stable
s2 d2 d0 0 …
If sign changes
s1 e1 0 0 …
Unstable s0 f0 0 0 …


 Example 8.2 Second-Order Case
A( s ) = s 2 + a1s + a0 If a1 & a0 > 0
Stable
Routh Array in the N=2 Case
s2 1 a0 If a1 > 0 & a0 < 0
s1 a1 0 or a1 < 0 & a0 < 0
1 pole in ORHP
s0 a0 0
a1a0 − (1)(0) If a1 < 0 & a0 > 0
bN − 2 = = a0
a1 Unstable


 Example 8.3 Third-Order Case

A( s ) = s + a2 s + a1s + a0
3 2

Routh Array in the N=3 Case
s3 1 a1 a0 a0
a1 − > 0 a1 >
s2 a2 a0 a2 a2
s1 a1-(a0/a2) 0 a0
a2 > 0, a1 > , a0 > 0
s0 a0 0 a2


 Example 8.4 Higher-Order Case
A( s ) = 6 s 5 + 5s 4 + 4s 3 + 3s 2 + 2 s + 1
Example 8.4
s5 6 4 2
s4 5 3 1
s3 0.4 0.8 0
s2 -7 1 0
s1 6/7 0 0
s0 1 0 0


 Example 8.5 Fourth-Order Case
A( s ) = s + s + 3s + 2 s + 2
4 3 2

Example 8.5
s4 1 3 2
s3 1 2 0
s2 1 2 0
s1 0 ≈ε 0 0
s0 2 0 0


Analysis of the Step Response
B(s) E (s) c
Y (s) = X ( s) Y ( s) = +
A( s ) A( s ) s
1 c = [ s Y ( s )]s =0 = H (0)
X (s) =
s  E (s) 
−1
y1 (t ) = L 
Y (s) =
B(s)  A( s ) 

A( s ) s y (t ) = y1 (t ) + H (0), t ≥ 0

Transient part Steady-state value (if stable)


 First-Order Systems
k k
H (s) = y (t ) = − (1 − e pt ), t ≥ 0
s− p p
1 k pt
X ( s) = y1 (t ) = e , t ≥ 0
s p
−k p k p k
Y (s) = + H (0) = −
s s− p p


 First-Order Systems : k
y (t ) = − (1 − e pt ), t ≥ 0
p

Without bound
p = 3, 2, 1
Unstable


 First-Order Systems : k
y (t ) = − (1 − e pt ), t ≥ 0
p

Bound Stable
p = -5, -2, -1
k = -p H(0) = 1

Steady-state value = 1


 First-Order Systems :
(time constant, τ)

≈ 63% of H(0)
p = -5, -2, -1
τ =0.2 sec

τ = 0.5 sec

τ = 1 sec


 Determining the pole location from the step
response

y(t) ≈ 1.73 ;
t ≈ 0.1 s

y (0.1) = 1.73 = 2[1 − e p ( 0.1) ] t = 0.1 sec
p = −20


 Second-Order Systems

k p1 = −ζω n + ωn ζ 2 − 1
H ( s) = 2
s + 2ζω n s + ωn2
p2 = −ζω n − ωn ζ 2 − 1

ζ is called the damping ratio

ωn is called the natural frequency


 Second-Order Systems :
Case when both poles are real

k k
H ( s) = y (t ) = (k1e p1t + k 2 e p2t + 1), t ≥ 0
( s − p1 )( s − p2 ) p1 p2
k
k ytr (t ) = (k1e p1t + k 2 e p2t ), t ≥ 0
Y ( s) = p1 p2
( s − p1 )( s − p2 ) s
k k
H ( 0) = 2 =
ωn p1 p2


k = 2, p1 = −1, p2 = −2 Case when both poles are real
2
H (s) =
( s + 1)( s + 2)
2
Y ( s) =
( s + 1)( s + 2) s
−2 1 1
Y ( s) = + +
s +1 s + 2 s
y (t ) = −2e −t + e − 2t + 1, t ≥ 0
ytr (t ) = −2e −t + e − 2t , t ≥ 0


Case when poles are real and repeated
k
H (s) =
( s + ωn ) 2
k
Y (s) =
( s + ωn ) 2 s

y (t ) =
k
[1 − (1 + ωnt )e−ω t ], t ≥ 0
n

ωn2
k
ytr (t ) = − (1 + ωnt )e −ω t , t ≥ 0
n

ωn2


Case when poles are real and repeated
k = 4, ωn = 2, p1 , p2 = −2
4
H (s) =
( s + 2) 2
4
Y ( s) =
( s + 2) 2 s
y (t ) = 1 − (1 + 2t )e − 2 t , t ≥ 0


Location of poles in the complex plane
p1 = −ζω n + jωd
p2 = −ζω n − jωd
k
H (s) =
( s − p1 )( s − p2 )
k
H (s) =
( s + ζω n − jωd )( s + ζω n + jωd )
k
H (s) =
( s 2 + 2ζω n + ωn ) + ωd
2 2


Case when poles are a complex pair
k k
H (s) = Y ( s) =
( s + ζω n ) 2 + ωd
2
( )
( s + ζω n ) 2 + ωd s
2

− ( k ωn ) s − 2kζ ωn k ωn
2 2
Y ( s) = +
( s + ζω n ) + ωd
2 2
s
− ( k ωn )( s + ζω n )
2
(kζ ωn ) k ωn2
= − +
( s + ζω n ) + ωd
2 2
( s + ζω n ) + ωd
2 2
s
k −ζω nt kζ −ζω nt k
y (t ) = − 2 e cos ωd t − e sin ωd t + 2 , t ≥ 0
ωn ωnωd ωn


k −ζω nt kζ −ζω nt k
y (t ) = − e cos ωd t − e sin ωd t + 2 , t ≥ 0
ωn2
ωnωd ωn

C cos β + D sin β = C 2 + D 2 sin( β + θ )
tan −1 (C D), C ≥ 0
where θ =
π + tan (C D), C < 0
−1

k  ωn −ζω nt 
y (t ) = 2 1 − e sin(ωd t + φ ), t ≥ 0
ωn  ωd 


17
H (s) =
s 2 + 2 s + 17
k = 17, ζ = 0.242, ωn = 17 , ωd = 4
p = −1 ± j 4
 17 −t 
y (t ) = 1 − e sin( 4t + 1.326), t ≥ 0
 4 


Effect of Damping Ratio on the Step Response
k
H (s) =
( s + ζω n ) + ωd
2 2

ωn = 1, k = 1
for ζ = 0.1, 0.25, 0.7


Effect of ωn on the Step Response
k
H (s) =
( s + ζω n ) 2 + ωd
2

ζ = 0.4, k = ωn 2

for ωn = 0.5, 1, 2 rad / s


Comparison of cases
k
H ( s) = 2 0<ζ<1 underdamped

ζ>1 overdamped

ζ=1 Critically damped


Comparison of cases
k
H (s) = 2
k = 4, ωn = 2
for ζ = 0.5, 1, 1.5


 Higher-Order Systems

bM s M + bM −1s M −1 +  + b1s + b0
H ( s) = N N −1
s + a N −1s +  + a1s + a0


Fourier Analysis of CT Systems
 LTI systems
Time domain Frequency domain

x(t) h(t) y(t) X(ω) H(ω) Y(ω)
F
h(t) is Impulse Response H(ω) is Frequency Response function
∞
y (t ) = h(t ) ∗ x(t ) = ∫ h(λ ) x(t − λ )dλ Y (ω ) = H (ω ) X (ω )
−∞

Assume that the system is stable: Amplitude : Y (ω ) = H (ω ) ⋅ X (ω )
∞
∫−∞
h(t ) dt < ∞ Phase : ∠Y (ω ) = ∠H (ω ) + ∠X (ω )


Response of LTI System to
Sinusoidal Inputs X(ω) H(ω) Y(ω)
F
x(t ) = A cos(ω0t + θ ) X (ω ) = Aπ [e − jθ δ (ω + ω0 ) + e jθ δ (ω − ω0 )]

Y (ω ) = H (ω ) X (ω )

Y (ω ) = AH (ω )π [e − jθ δ (ω + ω0 ) + e jθ δ (ω − ω0 )]
= Aπ [e − jθ H (−ω0 )δ (ω + ω0 ) + e jθ H (ω0 )δ (ω − ω0 )]
= Aπ H (ω0 ) [e − j (θ + ∠H (ω0 ))δ (ω + ω0 ) + e j (θ + ∠H (ω0 ))δ (ω − ω0 )]

F -1
Response to
y (t ) = A H (ω0 ) cos(ω0t + θ + ∠H (ω0 ))
Sinusoidal Input


Sinusoidal Inputs x(t) h(t) y(t)

x(t ) = A cos(ω0t + θ ) h(t) y (t ) = A H (ω0 ) cos(ω0t + θ + ∠H (ω0 ))

x(t ) = A1 cos(ω1t + θ1 ) + A2 cos(ω 2t + θ 2 ) h(t)

y (t ) = A1 H (ω1 ) cos(ω1t + θ1 + ∠H (ω1 )) + A2 H (ω2 ) cos(ω2t + θ 2 + ∠H (ω 2 ))

x(t ) = cos(100t ) + cos(3000t ) where ω1 = 100
h(t)
and ω2 = 3000
y (t ) = H (100) cos(100t + ∠H (100)) + H (3000) cos(3000t + ∠H (3000))


Sinusoidal Inputs X(ω) H(ω) Y(ω)

Y (ω ) = H (ω ) X (ω )
Vout (ω ) = H (ω )Vin (ω )

Vout (ω )
dvout (t ) H (ω ) =
RC + vout (t ) = vin (t ) Vin (ω )
dt
jω RCVout (ω ) + Vout (ω ) = Vin (ω ) Vout (ω ) 1
=
( jω RC + 1)Vout (ω ) = Vin (ω ) Vin (ω ) ( jω RC + 1)

1 1
H (ω ) = ; H (ω ) = ; ∠ H (ω ) = − tan −1 ωRC
( jω RC + 1) (ω RC ) 2 + 1


Sinusoidal Inputs

Low frequency
H(ω)

lim H (ω ) = 1
ω →0

High frequency
∠ H(ω)

lim H (ω ) = 0
ω →∞


Sinusoidal Inputs x(t) h(t) y(t)

x(t ) = cos(100t ) + cos(3000t ) h(t)

y (t ) = H (100) cos(100t + ∠H (100)) + H (3000) cos(3000t + ∠H (3000))

1
H (ω ) =
(ω RC ) 2 + 1
x(t ) = cos(100t ) + cos(3000t ) ∠ H (ω ) = − tan −1 ωRC

y (t ) = H (100) cos(100t + ∠H (100)) + H (3000) cos(3000t + ∠H (3000))

ω = ω 1 = 100 ω = ω2 = 3000


Sinusoidal Inputs
1
H (ω ) =
x(t ) = cos(100t ) + cos(3000t ) (ω RC ) 2 + 1
∠ H (ω ) = − tan −1 ωRC

y (t ) = H (100) cos(100t + ∠H (100)) + H (3000) cos(3000t + ∠H (3000))

RC = 0.001
H(ω)

x(t)
∠ H(ω)

y(t)


Sinusoidal Inputs
1
H (ω ) =
x(t ) = cos(100t ) + cos(3000t ) (ω RC ) 2 + 1
∠ H (ω ) = − tan −1 ωRC

y (t ) = H (100) cos(100t + ∠H (100)) + H (3000) cos(3000t + ∠H (3000))

RC = 0.01
H(ω)

x(t)
∠ H(ω)

y(t)


Periodic Inputs x(t) h(t) y(t)
∞
x (t ) = a0 + ∑ Ak cos(kω0t + θ k ), − ∞ < t < ∞ h(t)
k =1

∞
y (t ) = a0 H (0) + ∑ Ak H (kω0 ) cos(kω0t + θ k + ∠H ( kω0 )), − ∞ < t < ∞
k =1

1 x
A = A H (kω0 )
y x cky = Ak H (kω0 )
k k 2
θ ky = θ kx + ∠ H (kω0 ) ∠cky = θ kx + ∠ H (kω0 )

Akx , θ kx is the coefficients of the trigonometric FS for x(t)
Aky , θ ky is the coefficients of the trigonometric FS for y(t)


Periodic Inputs
 Response to a rectangular pulse train
∞
x(t)
x(t ) = a0 + ∑ Ak cos(kω0t + θ k ), − ∞ < t < ∞
… …
k =1

∞
2.0 -0.5 0.5 2.0 t x(t ) = a0 + ∑ ak cos(kπt ), − ∞ < t < ∞
k =1

 2
  , k = 1,3,5, 
0.5, k = 0 A =  kπ
x
k
 0,
 k = 2,4,6, 
ckx = 0, k = ±2,±4,±6, 
1 π , k = 3,7,11, 
 kπ , k = ±1,±3,±5,  θk = 
x
 0, all other k


Periodic Inputs
x(t)

… …
2.0 -0.5 0.5 2.0 t a0y = H (0)a0 = 0.5
x

ω = kω0  2 1
 , k = 1,3,5, 
2π Ak =  kπ (kπRC ) + 1
y 2
ω0 = ,T = 2 0,
T  k = 2,4,6, 
ω = kπ π − tan −1 kπRC , k = 3,7,11,
θ ky = 
− tan kπRC ,
−1
all other k


Periodic Inputs
• RC = 1


Periodic Inputs
• RC = 0.01


Nonperiodic Inputs
 Response to a rectangular pulse

X(ω) H(ω) Y(ω)

Y (ω ) = H (ω ) X (ω )
y (t ) = F - 1 { H (ω ) X (ω )}
1 ∞
y (t ) = ∫−∞ H (ω ) X (ω )e jωt dt
2π


Nonperiodic Inputs
x (t ) y (t ) = F - 1 {Y (ω )}
1
= F - 1 { H (ω ) X (ω )}
t 1 ∞
y (t ) = ∫ H (ω ) X (ω )e jωt dt
-1/2 1/2 2π −∞
F F-1
ω Y (ω ) = H (ω ) X (ω )
X (ω ) = sinc
2π H(ω)

1
H (ω ) =
jωRC + 1


Nonperiodic Inputs
• RC = 1

F-1
F

H(ω)


Nonperiodic Inputs
• RC = 0.1

F-1
F

H(ω)


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