Discrete Time Signal Correlation MCQ Quiz – Objective Question with Answer for Discrete Time Signal Correlation

Let us consider x(n) to be the input reference signal and y(n) to 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.

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,…

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).

We know that rxy(l)=\(\sum_{n=-\infty}^{\infty} x(n)y(n-l)\) where l=0,±1,±2,…

At l=0, rxy(0)=\(\sum_{n=-\infty}^{\infty} x(n)y(n)\)=7

For l>0, we simply shift y(n) to the right relative to x(n) by ‘l’ units, compute the product sequence and finally, sum over all the values of product sequence.

We get rxy(1)=13, rxy(2)=-18, rxy(3)=16, rxy(4)=-7, rxy(5)=5, rxy(6)=-3, rxy(l)=0 for l≥7

similarly for l<0, shift y(n) to left relative to x(n)

We get rxy(-1)=0, rxy(-2)=33, rxy(-3)=-14, rxy(-4)=36, rxy(-5)=19, rxy(-6)=-9, rxy(-7)=10, rxy(l)=0 for l≤-8
So, the sequence rxy(l)= {10,-9,19,36,-14,33,0,7,13,-18,16,-7,5,-3}

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)

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)

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)|≤\(\sqrt{r_{xx}(0).r_{yy}(0)}\).

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)=\(\frac{r_{xx}(l)}{r_{xx}(0)}\).

Let us consider a signal x(n) whose auto correlation is defined as rxx (l).
We know that for auto correlation sequence rxx (l)=rxx (-l).
So, auto correlation sequence is an even sequence.

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<∞)