EG1110 SIGNALS AND SYSTEMS
u(t)
y(t)
G
Input/Output analysis of systems
• An important question:
Consider a system
If we know
u(t)
G
y(t)
– the input signal; and
– a description of the system
Input: u(t)
How do we calculate the output?
Output: y(t)
• Often analysis of system behaviour done in time domain
i.e. observe output response y(t) when input u(t) is applied.
1
2
Answers...
Convolution - discrete time
• Static systems
• Convolution better demonstrated in discrete-time.
Simple:
• Discrete-time δ-function (impulse) central to discrete-time convolution:
System memoryless, so output at time t only depends on input at time t.





• Dynamics systems
δ(k − k0) = 

More complicated:


System has memory. so output at time t function of input at time t plus, typically, one or more
past inputs.
0
k 6= k0
1
k = k0
• Useful fact:
Fact 1: Any given discrete time signal can be represented as a sum of impulses.
• Convolution
A technique needed to find current output for a dynamic systems (principle simple, but calculation
can be difficult except in simple cases).
x(k) =
∞
X
x(n)δ(k − n)
n=−∞
Remember δ(k − n) is only non-zero when n = k
3
4
(1)
• Assume system is linear and time invariant.
Example:
Let x(k) be given by



















Thus x(n) is given by
2
k = −1
3
k=0
x(k) = 
3

k=1



































4
k=2
0
otherwise
• Now define
2
n = −1
3
n=0
x(n) = 
3

n=1
















4
n=2
0
otherwise
y(k) = Gk [u(k)]
Then using our summing fact we get

y(k) = Gk 
Thus we can calculate x(k) according to equation (1):
=
x(−2) =
etc. etc.
∞
X
x(−1) =
x(n)δ(−2 − n)
n=−∞
∞
X
=
x(n)δ(−1 − n)
n=−∞
= x(−2)δ(−2 + 2)
= x(−1)δ(−1 + 1)
= x(−2)
= x(−1)
= 0
= 2
∞
X
n=−∞
∞
X
n=−∞
∞
X
n=−∞

u(n)δ(k − n)
Gk [u(n)δ(k − n)]
by super − position
u(n)Gk [δ(k − n)]
by homogeneity and u(n) constant
Now define the time-shifted impulse response h(k − n) as
h(k − n) = Gk [δ(k − n)]
5
6
Thus we have
Example - discrete-time convolution
• Let impulse response of system be given by
y(k) =
∞
X
u(n)h(k − n)
n=−∞
h(k) = 3δ(k) + 2δ(k − 1)
i.e. The output of the system can be expressed in terms of its time-shifted impulse response summed over all time.
• Let input be
• This summation is called convolution










• Specifically:
y(k) = u(k) ? h(k) =
∞
X
n=−∞
u(n)h(k − n)
1
k=1
u(k) = 
3

k=2







0
otherwise
• Objective: calculate response of system using convolution sum.
• We know that
is called the convolution sum.
y(k) = u(k) ? h(k)
7
8
(2)
=
=
∞
X
n=−∞
∞
X
n=−∞
u(n)h(k − n)
u(n)[3δ(k − n) + 2δ(k − n − 1)]
y(2) = u(0)[3δ(2 − 0) + 2δ(2 − 0 − 1)]
(3)
(by using given impulse response)
• Input begins to be nonzero at k = 1. As system is causal, this is first time at which output can be
(n = 0)
+u(1)[3δ(2 − 1) + 2δ(2 − 1 − 1)]
(n = 1)
+u(2)[3δ(2 − 2) + 2δ(2 − 2 − 1)]
(n = 2)
= u(0)[0 + 0] + u(1)[0 + 2] + u(2)[3 + 0]
nonzero.....
= 0 + 2 + 9 = 11
...Hence
y(1) = u(0)[3δ(1 − 0) + 2δ(1 − 0 − 1)]
+u(1)[3δ(1 − 1) + 2δ(1 − 1 − 1)]
y(3) = u(0)[3δ(3 − 0) + 2δ(3 − 0 − 1)]
(n = 0)
(n = 1)
= u(0)[0 + 2] + u(1)[3 + 0]
= 0+3=3
(n = 0)
+u(1)[3δ(3 − 1) + 2δ(3 − 1 − 1)]
(n = 1)
+u(2)[3δ(3 − 2) + 2δ(3 − 2 − 1)]
(n = 2)
+u(3)[3δ(3 − 3) + 2δ(3 − 3 − 1)]
(n = 3)
= u(0)[0 + 0] + u(1)[0 + 0] + u(2)[0 + 2] + u(3)[3 + 0]
= 0+0+6+0=6
9
10
As impulse response only has one memory stage, output is zero after k = 3.










Hence we have computed output of system as















y(k) = 

3
k=1
11
k=2
6
k=3
0
otherwise












2
y2(k) = 
6








(4)
0
y(k) = y1(k) + y2(k)
i.e. the total response is the sum of the impulse responses - as expected!
Let h1(k) = 3δ(k) and h2(k) = 2δ(k − 1)
Then we would get outputs
3
k=1
y1(k) = 
9

k=2







0
11
k=3
otherwise
Obviously
• Note that system response to single impulses would be given as follows:










k=2
otherwise
12