Syracuse University: ELE 305: Quiz # 2 Solution: 2004.02.26.R
ELE 305: Discrete Time Signals and Systems
Quiz # 2 Solution
Total Questions = 4
Each question is worth 5 points
Total Points = 20
Open book, open notes
The sequence x[n]  [n  2]  [n  5] is input to a discrete time LTI system
Q1.
with an impulse response h[n]  [n]  [n  1]  [n  2]  [n  3]  [n  4]  [n  5] .
Determine the output, y[n].
There are a number of ways to arrive at the result of the convolution that needs to be carried out:: y[n] = x[n] * h[n]
The various signals involved are shown:
Q2.
With justifications, determine the causality and stability of the impulse
response
h[n]    n (sin( n)) n u[n] .
Causality: For a digital system to be causal, we require that the impulse function be equal to zero for n < 0. Since the
h[n] expression has a unit step function associated with it (which is zero for n < 0 and one for n > 0), this fits the definition
of causality. Hence system is causal
Stability: For a digital system to be stable, we require that the impulse response be absolutely summable.
h1[n]    n u[n] . We can show that this is absolutely summable, since it is a geometric


1
n
n

series with a ratio less than one. That is,   u[ n]  
. Therefore, a system with impulse
1   1
n  
n 0
Consider an impulse response,
response h1[n] would be stable.
For the impulse response h[n], note that sinn(n) is always bounded between +1. Therefore,


n  
n

sin n (n)   n
n 0
Since the right hand side converges to a finite value, the left hand side must converge to a finite value as well. Hence the
impulse response h[n] is absolutely summable and the system is stable.
Syracuse University: ELE 305: Quiz # 2 Solution: 2004.02.26.R
Q3.
The step response s[n] of a discrete time LTI system is given by
s[n]   n u[n] for 0    1
Find the impulse response h[n] of the system.
The relationship we are looking for is: h[n] = s[n] – s[n-1]. In this case, we get:
h[n]   n u[n]   n1u[n  1]
h[0] = 1
h[1] =  - 1
h[2] = ( - 1)
h[3] = ( - 1)
h[4] = ( - 1)
h[5] = ( - 1)
…
Summarizing, h[n] = [n] + n-1( - 1)u[n-1]
Q4.
A first approximation of the communication channel associated with
multipath propagation may be the following discrete time equation:
y[n]  x[n]  ax[n  1] , for 0  a  1
(that is, the output signal consists of the input signal plus a delayed and
scaled down version of the input signal). Could you find a causal inverse system
that would recover the original signal, x[n], from the output y[n]? Check
whether this inverse system is stable.
{Hint: Determine the impulse response of the communication channel, then try to
come up with an inverse impulse response such that the convolution of the two
ends up being [n ] .}
As shown in class, while going over the solutions for this quiz, the impulse response of the communication channel and
the inverse impulse response can be found to be as follows:
Communication Channel’s impulse response, h[n] = [n] + a [n-1]
The impulse response of the inverse system, h1[n] = (-a)n u[n]
For stability, we require to determine whether or not h1[n] is absolutely summable. Since h1[n] is a geometric series with
a ratio less than one, we know that it will converge to a finite number, that is:


n  
n 0
 (a) n u[n]  a n 
1
1 a