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Solving convolution problems
PART I: Using the convolution integral
The convolution integral is the best mathematical representation of the physical process that occurs when an input acts on a linear system to produce an output. If x ( t ) is the input, y ( t ) is the output, and h ( t ) is the unit impulse response of the system, then continuous-time convolution is shown by the following integral.
In it, τ is a dummy variable of integration, which disappears after the integral is evaluated.
Example 1: unit step input, unit step response Let x ( t ) = u ( t ) and h ( t ) = u ( t ).
The challenging thing about solving these convolution problems is setting the limits on t and τ. I usually start by setting limits on τ in terms of t , then using that information to set limits on t.
- The unit step function u ( τ ) makes the integrand zero for τ < 0, so the lower bound is 0.
- The unit step function u ( t − τ ) makes the integral zero for τ > t , so the upper bound is t.
- Once we have used the step functions to determine the limits, we can replace each step function with 1.
This integral produces y ( t ) = t. However, when we used t to set a limit on τ , we also created a limit on t. In this case, y ( t ) is zero when t < 0, because we have already set 0 < τ < t and there is no τ that satisfies 0 < τ < 0. Therefore, the answer to the convolution problem is
A system with h ( t ) = u ( t ) is known as an integrator, and can be made from an amplifier with a capacitor in it, or pretty much any system that accumulates an input and does not leak.
y t xt ht x τ ht τ d τ
+∞
−∞
y t u τ ut τ d τ
+∞
−∞
yt d τ
t
0
or () () 0 , 0
( ) yt tut t
tt y t =
Example 2: Unit step input, 1/ x response Let x ( t ) = u ( t ) and h ( t ) = u ( t )/( t +1). Convolution is commutative, so we can swap the t and t − τ and write the integral in either of these two ways.
The version on the left looks easier, so let’s try it.
- The unit step function u (τ) makes the integrand zero for τ < 0, so the lower limit is 0.
- The unit step function u ( t –τ) makes the integrand zero for τ > t , so the upper limit is t.
- Once we have used the step functions to determine the limits, we can replace each step function with 1.
This integral produces y ( t ) = ln( t +1). However, the fact that t is the upper limit on the range 0 < τ < t means that y ( t ) is zero when t < 0. Therefore, the solution is y ( t ) = ln( t +1) u ( t ).
Example 3: pulse input, unit step response. Let x ( t ) = u ( t ) – u ( t –2), h ( t ) = u ( t ).
The integrand is zero when τ < 0 and τ > 2, so that at most the integrand is non-zero when 0 < τ < 2. It is also zero when τ > t. This sets up three intervals for t. First, when t < 0 there is no way that 0 < τ < 2 and τ < t. Therefore, for t < 0, y ( t ) = 0. Next, when 0 < t < 2, the integrand is 1 when 0 < τ < t , making 0 and t the limits of integration. Finally, when t > 2, the value of t does not matter any more and the limits of integration are 0 and 2. Thus:
t < 0 y (^ t )=^0
0 < t < 2 yt d t
t
0
2 < t ()^12
2
0
yt = d τ =
τ τ d
t
y t ut u d u ut
+∞
−∞
+∞
−∞ − +
yt d
t
y t u τ u τ ut τ d τ
+∞
−∞
()= [ ( )− ( − 2 )] ( − )
Example 2: input is {1, 1 } pulse, response is {+1, –1 } Remember we are summing over k. The input is zero for k < 0, so the lower limit is 0. The system response is zero for k < n , so the upper limit is n , and the output is zero for n < 0. Therefore we start summing for n = 0.
n k =0 k =1 sum 0 x(0)h(0–0) = 1*1 = 1
x(1)h(0–1) =1*0 = 0
1 x(0)h(1–0) = 1*(–1) = –
x(1)h(1-1) =1*1=
2 x(0)h(2–0) = 10 = 0 x(1)h(2-1) = 1(–1) = –1 – 3 x(0)h(3–0) = 10 = 0 x(1)h(3-1) = 10 = 0 0
The result is that y ( n ) = { 1 0 –1 }, starting at n =0.
Problem 2. Find the output y ( n ) = x ( n )* h ( n ), where x ( n ) = { 1 , 1 } and h ( n ) = { 3 , 2, 1 }.
Both x ( n ) and h ( n ) are zero for n < 0, and the bold number shows where n = 0.
You may use either a table or the graphical method for finding y ( n ).
