Signal Processing question Is the system y[n] = x[n] - x[n-1] time invariant?

In summary, the system y[n] = x[n] - x[n-1] is time invariant since it does not explicitly depend on one of the inputs. Shifting the input and running it through the system will result in the same output as running the signal through the system first. This can be proven through a formal proof.
  • #1
seang
184
0

Homework Statement


Is the system y[n] = x[n] - x[n-1] time invariant?


Homework Equations





The Attempt at a Solution


I say no, but there's one thing I don't understand. I think if you shifted the input, and then ran those samples through the system, you'd get x[n-n0] - x[n-1].

If you ran the signal through the system first, then shifted the output, you'd get x[n-n0] - x[n-n0-1].

This is my thinking, is it incorrect?
 
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  • #2
seang said:

Homework Statement


Is the system y[n] = x[n] - x[n-1] time invariant?


Homework Equations





The Attempt at a Solution


I say no, but there's one thing I don't understand. I think if you shifted the input, and then ran those samples through the system, you'd get x[n-n0] - x[n-1].

If you ran the signal through the system first, then shifted the output, you'd get x[n-n0] - x[n-n0-1].

This is my thinking, is it incorrect?

As for the answer, the system is time invariant since the system does not explicitly depend on one of the inputs.

As for your method of approaching this, even if you shift the input and run the samples through the system, you'd get the same output as running the signal through the system first.

Here's a formal proof you can look at to see why it is true.

http://en.wikipedia.org/wiki/Time-invariant
 
  • #3
146kok is correct.

Your error comes in the first answer where you shifted the input and then ran it through the equation. What the equation says is that y is equal to the present input plus the input just before the present input. If the present input is x(n-n0) then the input just before the present input has to be x(n-n0-1). This agrees with your second answer so it is time invariant.
 

1. What is signal processing and why is it important?

Signal processing is a field of study that deals with analyzing and manipulating signals, which are data that vary over time. It is important because signals are used to represent information in many different applications, such as communication systems, biomedical devices, and image and audio processing.

2. What are the different types of signal processing?

There are two main types of signal processing: analog and digital. Analog signal processing involves manipulating signals that are continuous in time, while digital signal processing involves manipulating signals that are represented by a sequence of numbers.

3. What are some common techniques used in signal processing?

Some common techniques used in signal processing include filtering, modulation, and Fourier analysis. Filtering is used to remove unwanted noise from a signal, while modulation is used to convert a signal from one form to another. Fourier analysis is used to decompose a signal into its individual frequency components.

4. What is the difference between time-domain and frequency-domain signal processing?

In time-domain signal processing, signals are analyzed and manipulated in the time domain, which means that the values of the signal are plotted against time. In frequency-domain signal processing, signals are analyzed and manipulated in the frequency domain, which means that the signal is decomposed into its individual frequency components.

5. How is signal processing used in real-world applications?

Signal processing is used in a wide range of real-world applications, such as audio and video compression, wireless communication, medical imaging, and radar and sonar systems. It is also used in everyday devices like smartphones, TVs, and music players to improve the quality and efficiency of the signals being processed.

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