Qualcomm Digital Signal Processing Quiz-1

For a discrete time signal, the value of x(n) exists only at integral values of n. So, for a non- integer value of ‘n’ the value of x(n) does not exist.

When we plot the graph for the given function, we get a straight line passing through origin with a unit positive slope. So, the function is called as unit ramp signal.

Given x(n)=aⁿ=(r.e^jθ)ⁿ =rⁿ.e^jnθ
=>x(n)=rⁿ.(cosnθ+jsinnθ)
Phase function is tan^-1(cosnθ/sinnθ)=tan^-1(tan nθ)=nθ.

We have used the magnitude-squared values of x(n), so that our definition applies to complex-valued as well as real-valued signals. If the energy of the signal is finite i.e., 0<E<∞ then the given signal is known as Energy signal.

The given signal is defined only when n=k by the definition of delta function. So, x(n)*δ(n-k)= x(k)*δ(n-k).

According to the definition of anti-symmetric signal, the signal x(n) should be symmetric over origin. So, for the signal x(n) to be symmetric, it should satisfy the condition x(n)=-x(-n).

Let x(t)=xe(t)+xo(t)
=>x(-t)=xe(-t)-xo(-t)
By subtracting the above two equations, we get
xo(t)=(1/2)*(x(t)-x(-t)).

If the signal x(n) was originally obtained by sampling a signal xa(t), then x(n)=xa(nT). Now, y(n)=x(2n)(say)=xa(2nT). Hence the time scaling operation is equivalent to changing the sampling rate from 1/T to 1/2T, that is to decrease the rate by a factor of 2. So, time scaling is also called as down-sampling.

When the value of ‘r’ lies between 0 and 1 then the value of x(n) goes on decreasing exponentially with increase in value of ‘n’. So, the signal is called as decaying exponential signal.

According to the definition of the impulse function, it is defined only at n=0 and is not defined elsewhere which is as per the signal given.