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Signals and Systems - Linear Time Invariant Discrete Time Systems - Sample Problems and Important Short Answers: Linear Time Invariant Discrete Time Systems

**1. **States the properties of
convolution

** **i).Commutative
property of convolution

**x(n)** ^{∗}**h(n)=h(n)** ^{∗}**x(n)=y(n**)

ii).Associative
property of convolution

**[x(n)** ^{∗}**h _{1}(n)**

iii).Distributive
property of convolution

**x(n)** ^{∗}**[ ****h _{1}(n)**

**2.** **Define non recursive and recursive of the
following system.**

When the
output y(n) of the system depends upon present and past inputs then it is
called non-recursive system. When the output y(n) of the system depends upon
present and past inputs as well as past outputs, then it is called recursive
system.

**3.
****Define
convolution sum?**

If x(n)
and h(n) are discrete variable functions, then its convolution sum

y(n) is
given by,

y(n)=_
x(k) h(n-k)

**4. If x(n) and y(n) are discrete variable
functions, what is its convolution sum. **

The
convolution sum is,

**5.
****Determine
the system function of the discrete time system described by the difference
equation.**

**Y(n) = 0.5y(n-1)+x(n)**

Taking
z-transform of both sides,

Y(z) =
0.5z^{-1}Y(z)+X(z)

H(z) =
Y(z)/X(z) = 1/(1 – 0.5z ^{-1})

**6.
****A causal
LTI system has impulse response h(n), for which the z-transform is H(z) = (1+z ^{-1})/(1-0.5z^{-1})(1+0.25z^{-1}).
Is the system stable? Explain.**

H(z) can
be written in terms of positive powers of z as follows:

H(z) =
z(z+1)/(z-0.5)(z+0.25)

Poles are
at p_{1} =0.5 and p_{2} = -0.25. Since both the poles are
inside unit circle. This system is stable.

**7.** **Check whether the system with system
function H(Z) = (1/1-0.5z ^{-1})+(1/1-2z^{-1}) with ROC |z| <
0.5 is causal and stable?**

H(z) =
z/(z – 0.5) + z/(z - 2). Poles of this sys tem are located at z = 0.5 and z =
2. This system is not causal and stable, since all poles are not located inside
unit circle.

**8.** **Is the discrete time system described by
the difference equation y(n) = x(-n) is causal?**

Here
y(-2) = x(-(-2)) = x(2). This means output at n=-2 depends upon future inputs.
Hence this system is not causal.

**9. Consider a system whose impulse is h(t) = e ^{-|t|}.
Is this system is causal or non causal? **

Here h(t)
= e^{-|t|}

=e^{-t} for t>=0

=e^{t} for t<0

Since
h(t) is not equal to zero for t<0, the system is non causal.

**10. Find the step response of the system if the
impulse response**

**10.
****Obtain
the convolution of**

x(n)*[h_{1}(n)+h_{2}(2)]
= x(n)*h_{1}(n) + x(n)*h_{2}(n)

**12.
****List the
steps involved in finding convolution sum? **

o folding

o
Shifting

o Multiplication
o Summation

**13.
Consider an LTI system with impulse response h(n)= δ(n-* _{“}) for an
input x(n), find the Y(e^{j}^{ω}**

Here is
the spectrum of output. By convolution theorem we can write,

**14.
****List the
properties of convolution?**

o
Commutative property of convolution

x(n) *
h(n) = h(n) * x(n) = y(n)

o
Associative property of convolution

[ x(n) *
h_{1}(n)] * h_{2}(n) = x(n) * [h_{1}(n) * h_{2}(n)]

o
Distributive property of convolution

x(n) * [h_{1}(n)
+ h_{2}(n)] = x(n) * h_{1}(n) + x(n) * h_{2}(n)

**15.
****Define
system function?**

H(z)=
Y(z) is called system function.It is the z transform of the unit sample X(Z)
response h(n) of the system.

**Sample Problems:**

**1. Consider the system described by the difference equation.**

**2. Given y[-1]=1 and y[-2]=0. Compute recursively a few terms of the following 2nd order DE:**

**3. Compute the impulse response of the system described by,**

**4. Obtain the structures realization of LTI system**

**5. Find the convolution of x(n)=[1,1,1,1,2,2,2,2] with h(n)=[3,3,0,0,0,0,3,3] by using matrix method.**

**Solution: By using matrix method, N=8**

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Signals and Systems : Linear Time Invariant Discrete Time Systems : Sample Problems and Important Short Answers: Linear Time Invariant Discrete Time Systems |

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