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I was reading in a book that below stated are the Properties of Discrete-time Sinusoidal Signal:

Property 1. The discrete time sinusoids whose frequency are separated by an integer multiple of 2$\pi$ are identical.

Property 2. The frequency of oscillation of discrete time sinusoids sequence increases as $\omega$ increases from 0 to $\pi$. If $\omega$ is increased from $\pi$ to 2$\pi$ then frequency of oscillation decreases.

I was able to understand the mathematical property implementation of above two property but have some basic questions for clarification:

  1. Is it that the above properties are applicable for all Discrete-time periodic Signals?
  2. As per property 1 - if two sinusoids signal has frequency separated by integer multiple of 2$\pi$ - then the signal having the greater frequency will have more cycles than other. Then how they both can be same?
  3. As per property 2 - what is meant by "frequency of oscillation"? How can a frequency of a signal vary (increase) from 0 to $\pi$ and (decrease) from $\pi$ to 2$\pi$ - isn't it should be same from 0 to $2\pi$?
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Any discrete-time periodic signal with period $N$ can be written in terms of its DFT coefficients $X_k$:

$$x(n)=\frac{1}{N}\sum_{k=0}^{N-1}X_ke^{j(2\pi/N) nk}$$

So whatever is true for a single sinusoid (or complex exponential) remains true for a general periodic signal.

Two discrete-time sinusoids with a frequency separation of $2\pi k$ are identical because the sine function is periodic with period $2\pi$:

$$x(n)=\sin(\omega n)\\ y(n)=\sin[(\omega+2\pi k)n]=\sin(\omega n +2\pi k n)=\sin(\omega n)=x(n)$$

Finally, the maximum frequency for a discrete-time sinusoid (and consequently for any discrete-time signal) is $\omega=\pi$:

$$\cos(\pi n)=(-1)^n$$

If the frequency is further increased you will observe aliasing:

$$\cos[(\pi+\Delta\omega)n]=\cos[(\pi+\Delta\omega)n-2\pi n]=\cos[(\Delta\omega-\pi)n]=\\=\cos[(\pi-\Delta\omega)n],\quad 0\le\Delta\omega\le\pi$$

So from the maximum frequency of $\pi$ you get a decrease in frequency down to frequency $0$ (for $\Delta\omega=\pi$).

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  • $\begingroup$ Thanks for all the details. However can you please explain how : sin[(π+Δω)n]=sin[(π+Δω)n−2πn] and isn't The sine of 180 degrees or π is 0 - so how $sin(πn)=(−1)^{n}$ ? $\endgroup$ – Programmer Aug 12 '14 at 17:13
  • $\begingroup$ @Prakash: Because $\sin(x)=\sin(x\pm 2\pi n)$. In this case $x=(\pi+\Delta\omega)n$. $\endgroup$ – Matt L. Aug 12 '14 at 17:18
  • $\begingroup$ Thanks again one last question - I have modified my comment regarding sin(πn)? $\endgroup$ – Programmer Aug 12 '14 at 17:19
  • $\begingroup$ Thanks for for the updates, I am able to clarify al my doubts $\endgroup$ – Programmer Aug 12 '14 at 17:22
  • $\begingroup$ @Prakash: Sorry for the mistake, I've edited my answer. Should have used cosine instead of sine. $\endgroup$ – Matt L. Aug 12 '14 at 17:22

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