7ECE Mid-term Attempt any two Subject: - DSP Marks:10
Q.1 Explain with mathematical equations and block diagram the concept of continuous time processing of discrete time signal. [5]
Ans.: Continuous time processing of discrete time signal: In this process, the input and output are in discrete form whereas the intermediate signals are in continuous in nature.
The output of the D/C converter is given as,
Xc(f) =
Q.2 A] The discrete time systems are represented by difference equation
a) y(n) = 3 y2 (n-1) – n x(n) + 4 x(n-1) - 2x(n+1) b) y(n) = x(n+1) - 3 x(n) + x(n-1); n ≥ 0
Are these systems linear, shift Invariant and Causal [2.5]
Ans:--
- Causal
a)y(n) = 3 y2 (n-1) – n x(n) + 4 x(n-1) - 2x(n+1)
Hence y(n) is determined using the past input sample value 4x(n-1) and future input sample value 2x(n+1) and hence system is non-causal.
b)y(n) = x(n+1) - 3 x(n) + x(n-1); n ≥ 0
Hence y(n) is determined using the past input sample value x(n-1) and future input sample value x(n+1) and hence system is non-causal.
- Linear
a)y(n) = 3 y2 (n-1) – n x(n) + 4 x(n-1) - 2x(n+1)
If we apply two inputs x1 (n)andx2 (n) separately, then the outputs become
y1 (n)= T {x1(n)} = 3 y2 (n-1) – n x1(n) + 4 x1(n-1) - 2x1(n+1)
y2 (n)= T {x2(n)} = 3 y2 (n-1) – n x2(n) + 4 x2(n-1) - 2x2(n+1)
Response of the system to linear combination of inputs x1 (n)andx2 (n) will be,
y3 (n)= T {a1 x1(n)+ a2 x2 (n)}
= a1 [3 y2 (n-1) – n x1(n) + 4 x1(n-1) - 2x1(n+1)] + a2 [3 y2 (n-1) – n x2(n) + 4 x2(n-1) - 2x2(n+1)]
The linear combination of two outputs given by will be:
y'3(n) = a1 y1(n)+ a2 y2 (n)
On comparing both the equations we observe that,
y3 (n) ǂ y'3 (n), hence the system is nonlinear.
b)y(n) = x(n+1) - 3 x(n) + x(n-1); n ≥ 0
If we apply two inputs x1 (n)andx2 (n) separately, then the outputs become
y1 (n)=T {x1(n)} = x1(n+1) – 3x1(n) + x1(n-1)
y2 (n)= T {x2(n)} = x2(n+1) – 3x2(n) + x2(n-1)
Response of the system to linear combination of inputs x1 (n)andx2 (n) will be,
y3 (n)= T {a1 x1(n)+ a2 x2 (n)}
= a1 [x1(n+1) – 3x1(n) + x1(n-1)] + a2 [x2(n+1) – 3x2(n) + x2(n-1)]
The linear combination of two outputs given by will be:
y'3(n) = a1 y1(n)+ a2 y2 (n) = a1 [x1(n+1) – 3x1(n) + x1(n-1)] + a2 [x2(n+1) – 3x2(n) + x2(n-1)]
on comparing both the equations we observe that,
y3 (n) = y'3 (n), hence the system is linear.
iii] Shift Invariant
a] Shift invariant
b] Shift invariant
B] Explain Linear systems with Linear phase. [2.5]
Ans.: A system is referred to as a generalized-linear-phase system if
H(ejω) = A(ejω)e−jαω+jβ ------(1)
A(ejω) is real function of ω. The phase is
θ(ω) = −αω+β
Note the phase could be called affine phase since the phase is an affine transformation of ω.
The group delay is
grp[H(ejω)] = -dθ(ω) dω
grp[H(ejω)] = α
if 2α is an integer, we might have impulse response symmetry about α.
Sinceand / H(ejω) = A(ejω)[cos(β−ωα)+ jsin(β−ωα)]
H
n=−∞n=−∞
we have
Therefore
∑
n=−∞
That is constant group delay implies Equation 2, so Equation .2 is a necessary condition for generalized linear phase. However, Equation 2 is not a sufficient condition for generalized linear phase.
As examples, two types of generalized linear phase systems are given below:
1.The type of systems is defined by
β = 0 or π
2α = M = an integer h[2α−n] = h[n]
It is easy to show
∑
n=−∞
2.The type of systems is defined by
2α = M = an integer h[2α−n] = −h[n]
It is easy to show
n
Q.3 Determine Frequency response and plot Magnitude and phase response of equation given by
y(n) - y(n-1) + 3/16 y(n-2) = x(n) - 1/2 x(n-1)
Ans.: For the impulse response, x(n)=δ(n) and hence y(n)=h(n). Therefore, the given function becomes
h(n) - h(n-1) + 3/16 h(n-2) = δ(n) - 1/2 δ(n-1)
Taking z-transformation, we get
H(z) – z-1 H(z) + 3/16 z-2 H(z) = 1 - 1/2 z-1
Therefore, H(z) = 1 - 1/2 z-1
1-z-1 +3/16 z-2
= z (z- ½)
z2–z+3/16
= z (z- ½)
(z- ¼) (z- ¾)
Now plot magnitude and phase plot by varying angular frequency w.