## Digital Signal Processing Questions and Answers Part-4

1. Which of the following parameters are required to calculate the correlation between the signals x(n) and y(n)?
a) Time delay
b) Attenuation factor
c) Noise signal
d) All of the mentioned

Answer: d
Explanation: Let us consider x(n) be the input reference signal and y(n) be the reflected signal.
Now, the relation between the two signals is given as y(n)=αx(n-D)+w(n)
Where α-attenuation factor representing the signal loss in the round-trip transmission of the signal x(n)
D-time delay between the time of projection of signal and the reflected back signal
w(n)-noise signal generated in the electronic parts in the front end of the receiver.

2. The cross correlation of two real finite energy sequences x(n) and y(n) is given as __________
a) $$r_{xy}(l)=\sum_{n=-\infty}^{\infty}x(n)y(n-l)$$ where l=0,±1,±2,…
b) $$r_{xy}(l)=\sum_{n=0}^{\infty}x(n)y(n-l)$$ where l=0,±1,±2,…
c) $$r_{xy}(l)=\sum_{n=-\infty}^{\infty}x(n)y(n-l)$$ where -∞<l<∞
d) none of the mentioned

Answer: a
Explanation: If any two signals x(n) and y(n) are real and finite energy signals, then the correlation between the two signals is known as cross correlation and its equation is given as
rxy(l)=$$\sum_{n=-\infty}^{\infty} x(n)y(n-l)$$ where l=0,±1,±2,…

3. Which of the following relation is true?
a) rxy(l)= rxy(-l)
b) rxy(l)= ryx(l)
c) rxy(l)= ryx(-l)
d) none of the mentioned

Answer: c
Explanation: we know that, the correlation of two signals x(n) and y(n) is
$$r_{xy}(l)=\sum_{n=-\infty}^{\infty} x(n)y(n-l)$$
If we change the roles of x(n) and y(n),
we get $$r_{yx}(l)=\sum_{n=-\infty}^{\infty} y(n)x(n-l)$$
Which is equivalent to
$$r_{yx}(l)=\sum_{n=-\infty}^{\infty} x(n)y(n+l)$$ => $$r_{yx}(-l)=\sum_{n=-\infty}^{\infty} x(n)y(n-l)$$
Therefore, we get rxy(l)= ryx(-l).

4. Which of the following relation is true?
a) ryx(l)=h(l)*ryy(l)
b) rxy(l)=h(l)*rxx(l)
c) ryx(l)=h(l)*rxx(l)
d) none of the mentioned

Answer: c
Explanation: Let x(n) be the input signal and y(n) be the output signal with impulse response h(n).
We know that y(n)=x(n)*h(n)=$$\sum_{k=-\infty}^{\infty} x(k)h(n-k)$$
The cross correlation between the input signal and output signal is
ryx(l)=y(l)*x(-l)=h(l)*[x(l)*x(-l)]=h(l)*rxx(l).

5. Which of the following is the auto correlation of x(n)?
a) rxy(l)=x(l)*x(-l)
b) rxy(l)=x(l)*x(l)
c) rxy(l)=x(l)+x(-l)
d) None of the mentioned

Answer: a
Explanation: We know that, the correlation of two signals x(n) and y(n) is rxy(l)=$$\sum_{n=-\infty}^{\infty} x(n)y(n-1)$$
Let x(n)=y(n) => rxx(l)=$$\sum_{n=-\infty}^{\infty}x(n)x(n-l)$$ = x(l)*x(-l)

6. What is the energy sequence of the signal ax(n)+by(n-l) in terms of cross correlation and auto correlation sequences?
a) a2rxx(0)+b2ryy(0)+2abrxy(0)
b) a2rxx(0)+b2ryy(0)-2abrxy(0)
c) a2rxx(0)+b2ryy(0)+2abrxy(1)
d) a2rxx(0)+b2ryy(0)-2abrxy(1)

Answer: c
Explanation: The energy signal of the signal ax(n)+by(n-l) is
$$\sum_{n=-\infty}^{\infty}[ax(n)+by(n-l)]^2$$
= $$a^2 \sum_{n=-\infty}^{\infty}x^2(n)+b^2 \sum_{n=-\infty}^{\infty}y^2(n-l)+2ab \sum_{n=-\infty}^{\infty}x(n)y(n-l)$$
= a2rxx(0)+b2ryy(0)+2abrxy(l)

7. What is the relation between cross correlation and auto correlation?
a) |rxy(l)|=$$\sqrt{r_{xx}(0).r_{yy}(0)}$$
b) |rxy(l)|≥$$\sqrt{r_{xx}(0).r_{yy}(0)}$$
c) |rxy(l)|≠$$\sqrt{r_{xx}(0).r_{yy}(0)}$$
d) |rxy(l)|≤$$\sqrt{r_{xx}(0).r_{yy}(0)}$$

Answer: d
Explanation: We know that, a2rxx(0)+b2ryy(0)+2abrxy(l) ≥0
=> (a/b)2rxx(0)+ryy(0)+2(a/b)rxy(l) ≥0
Since the quadratic is nonnegative, it follows that the discriminate of this quadratic must be non positive, that is 4[r2xy(l)- rxx(0) ryy(0)] ≤0 => |rxy(l)|≤$$\sqrt{r_{xx}(0).r_{yy}(0)}$$.

8. The normalized auto correlation ρxx(l) is defined as _____________
a) $$\frac{r_{xx}(l)}{r_{xx}(0)}$$
b) –$$\frac{r_{xx}(l)}{r_{xx}(0)}$$
c) $$\frac{r_{xx}(l)}{r_{xy}(0)}$$
d) None of the mentioned

Answer: a
Explanation:If the signal involved in auto correlation is scaled, the shape of auto correlation does not change, only the amplitudes of auto correlation sequence are scaled accordingly. Since scaling is unimportant, it is often desirable, in practice, to normalize the auto correlation sequence to the range from -1 to 1. In the case of auto correlation sequence, we can simply divide by rxx (0). Thus the normalized auto correlation sequence is defined as ρxx(l)=$$\frac{r_{xx}(l)}{r_{xx}(0)}$$.

9. What is the auto correlation of the sequence x(n)=anu(n), 0<a<l?
a) $$\frac{1}{1-a^2}$$ al (l≥0)
b) $$\frac{1}{1-a^2}$$ a-l (l<0)
c) $$\frac{1}{1-a^2}$$ a|l|(-∞<l<∞)
d) All of the mentioned

Answer: d
Explanation:
rxx(l)=$$\sum_{n=-\infty}^{\infty} x(n)x(n-l)$$
For l≥0, rxx(l)=$$\sum_{n=l}^{\infty} x(n)x(n-l)$$
=$$\sum_{n=l}^{\infty} a^n a^{n-l}$$
=$$a^{-l}\sum_{n=l}^{\infty} a^{2n}$$
=$$\frac{1}{1-a^2}a^l$$(l≥0)
For l<0, rxx(l)=$$\sum_{n=0}^{\infty} x(n)x(n-l)$$
=$$\sum_{n=0}^\infty a^n a^{n-l}$$
=$$a^{-l}\sum_{n=0}^{\infty} a^{2n}$$
=$$\frac{1}{1-a^2}a^{-l}$$
So, rxx(l)=$$\frac{1}{1-a^2}a^{|l|}$$ (-∞<l<∞)

10. If x(n) is the input signal of a system with impulse response h(n) and y(n) is the output signal, then the auto correlation of the signal y(n) is?
a) rxx(l)*rhh(l)
b) rhh(l)*rxx(l)
c) rxy(l)*rhh(l)
d) ryx(l)*rhh(l)

Answer: b
Explanation: ryy(l)=y(l)*y(-l)
=[h(l)*x(l)]*[h(-l)*x(-l)]
=[h(l)*h(-l)]*[x(l)*x(-l)]
=rhh(l)*rxx(l).