In engineering applications, we frequently encounter functions whosevalues change abruptly at specified values of time t. One commonexample is when a voltage is switched on or off in an electrical circuitat a specified value of time t.Shifted unit step functionRectangular pulseThe value of t = 0 is usually taken as a convenient time to switch on oroff the given voltage.The switching process can be described mathematically by thefunction called the Unit Step Function (otherwise known as theHeaviside function after Oliver Heaviside).The Unit Step FunctionDefinition: The unit step function, u(t), is defined asThat is, u is a function of time t, and u has value zero when time isnegative (before we flip the switch); and value one when time ispositive (from when we flip the switch).
Shifted Unit Step FunctionIn many circuits, waveforms are applied at specified intervals otherthan t = 0. Such a function may be described using the shifted (akadelayed) unit step function.Definition of Shifted Unit Step Function:A function which has value 0 up to the time t = a and thereafter hasvalue 1, is written:Example of Shifted Unit Step Function:f(t) = u(t − 3)The equation means f(t) has value of 0 when t < 3 and 1 when t > 3.The sketch of the waveform is as follows:
Rectangular PulseA common situation in a circuit is for a voltage to be applied at aparticular time (say t = a) and removed later, at t = b (say). Such asituation is written using unit step functions as:V(t) = u(t − a) − u(t − b)This voltage has strength 1, duration (b − a).EXAMLPEThe graph of V(t) = u(t − 1.2) − u(t − 3.8) is as follows. Here, theduration is 3.8 − 1.2 = 2.6.EXERCISES
Write the following functions in terms of unit stepfunction(s). Sketch each waveform.(a) A 12-V source is switched on at t = 4 s.AnswerSince the voltage is turned on at t = 4, we need to use u(t − 4). Wemultiply by 12 since that is the voltage.
(b)(Assume a > 0.)AnswerIn words, the voltage has value 1 up until time t = a. Then it is turnedoff.We have a "rectangular pulse" situation and need to use this formula:V(t) = u(t − a) − u(t − b)In our example, the pulse starts at t = 0 [so we use u(t)] and finishesat t = a [so we use u(t − a)].So the required function is:
(c) One cycle of a square wave, f(0) = 4, amplitude = 4, period = 2seconds.Answerf(0) = 4 means we start at value 4.If the whole wave has period 2, and it is a square wave, then it meansfor half of the time, the value is (positive) 4 and the other half it isnegative 4.So for the first second, it has value 4, for the second second, thefunction value is -4.We write this, using the "rectangular pulse" formula from before:The graph of this first cycle is:
(d) The unit Ramp function (i.e. f(t) = t for t > 0)AnswerThe unit ramp function has slope 1 [so the function is simply V(t) = t],starting from t = 0 [so we need to multiply by u(t)], and passesthrough (0, 0).So the voltage function is given by:V(t) = t • u(t)The graph of the function is:
(e) One cycle of a sawtooth waveform (i.e. for 0 < t < b.Assume a > 0.)AnswerOur graph starts at t = 0 and has slope a / b. It finishes at t = b.So our function will be:The graph of our function:
(f)AnswerIn this example, our function is V(t) = 2t + 8 which has slope 2 and V-intercept 8.The signal is only turned on between t = 3 and t = 5. The rest of thetime it is off.So our voltage function will be:The graph is as follows: