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Friday
Dec302011

Convolution in DSP

One of the problems that I've encountered in trying to teach myself DSP has been that most textbooks contain questions with no answers. Having been booked to take part in a 3 day course at Oxford Uni that was cancelled due to insufficient take-up, I've picked up the textbook that they were to use. It's a great book and has self assessment questions throughout with answers in the back, but the end of chapter exams don't. Some have a numerical answer which is fine for me to check I'm getting the right end of the stick, but not much help if I'm completely lost. Hopefully someone who knows what they're talking about will take pity on me and let me know if I've got this one right or not...

Given a signal x(t) and an impulse response h(t) as shown below, determine the output of the system y(t).

At first I was completely lost trying to do it purely mathematically, so I followed the example in the book and visualised the problem by splitting it into different regions and using the convolution integral for each.

The first region, t<0, is simple enough as x(t) = 0, therefore x(t)h(t-τ) is also 0, thus y(t) = 0.

The second region, 0<t<1, involves the product of x(t) with h(t-τ) which can be visualised as below.

Resulting in the following x(t)h(t-τ) product.

Taking the integral of this product gives us y(t) for 0<t<1.

Region 3 covers 1<t<2, where x(t) is holding a constant 4 and h(t-τ) is fully within the region of x(t)>0, hence y(t) for this range can be found from the convolution integral as:

Region 4, 2<t<3 contains the remaining part of h(t-τ) remaining within the time that x(t) is >0, visualised as:

Mathematically, this gives us:

Finally, region 5 consists the remainder of the signal where 3<t. In this region, y(t) = 0.

When combined, this gives a combined output from this system as illustrated in this final chart.

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