GATE Bits - Mathematical Representation of Signals
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1 Bank PO - Bank Clerk - GATE Insurance - SSC CGL - BSNL TTA - RBI Testbook Blog - Testbook Mobile App - Daily Current Affairs GK Quiz Mobile App - by Testbook.com GATE Bits - Mathematical Representation of Signals It is a very effective ability to be able to represent signals in mathematical form. In this blog, we learn the mathematical representation of signals in terms of basic functions like impulse, unit step and ramp function. We take examples to learn it. This GATE Bit will help you crack GATE EC 2016 or GATE EE On a general note, it should be learnt that whenever there is an amplitude change only at a single point, there is an impulse appearing at that point. If the amplitude changes for a range of points beyond that particular point, a step signal is involved. If the slope of the signal changes across a point then a ramp signal appears at that point. Determining Amplitude of Signals Impulse Signal: If the instantaneous change in amplitude is positive then a positive impulse is involved. If the amplitude change is negative, then a negative impulse is involved. The magnitude of impulse is determined by the difference in magnitude change. Step Signal: 1 / 6
2 A step amplitude is determined by subtracting amplitude just before the transition from the amplitude just after the transition. Ramp Signal: A ramp amplitude is determined by subtracting initial slope from final slope. Examples for Mathematical Representation of Signals Example 1: Step Signal As we can see that the amplitude changes for points after the origin, we know that a step signal appears at origin. The magnitude of step is determined by subtracting amplitude in section a from amplitude b we have b a = 1. Thus, a unit step u(t) appears at origin. Similarly, subtracting amplitudes across t o we get c b = 0 1 = 1. Thus a unit step u(t t o ) appears at t o. Thus, x(t) = u(t) u(t t o ). Example 2: Ramp Signal & Step Signal 2 / 6
3 We see the slope of section a is 1, so a ramp r(t) appears at the origin. There is an amplitude shift of 2 at t = 1. Thus, there is a step signal of magnitude 2 appearing at t = 1. Thus, a signal is 2u(t 1). Now, since there is no slope change across t = 1 (slope = 1 after and before t = 1), there is no ramp signal involved at t = 1, though it seems like another ramp is involved at the point. Thus x 2 (t) = r(t) + 2u(t 1) Example 3: Ramp Signal & Impulse Signal 3 / 6
4 Slope is section a = 3/1 = 3. Thus, a ramp of slope 3 appears at origin. We have 3r(t). Change is slope across t = 1 is 0 3 = 3. So, a ramp of slope 3 appears at t = 1. We have 3r(t 1). Also, there is an amplitude change of 4 3 = 1 at point t = 1 alone. So, we have an impulse of magnitude 1 appearing at t = 1. We have?(t 1). Thus x 2 (t) = 3r(t) +?(t 1) 3r(t 1). Example 4: Mixed Signal of Impulse, Step & Ramp Signals 4 / 6
5 Across origin slope changes by 2 0 = 2. So, a ramp of slope 2 appears at origin. At t = 1, there are three signals appearing. One impulse of magnitude 6 4 = 2 due to amplitude change appearing only at point t = 1. Step input due to amplitude change 4 2 = 2 at point t = 1. Finally, a ramp due to slope change of 0 (in section b) 2 (in section a) = 2. Finally, at point t = 3 a step input appear due to amplitude change of 0 4 = 4. Thus, x 3 (t) = 2r(t) + 2?(t 1) + 2u(t 1) 2r(t 1) 4u(t 3) Try it Yourself: Take Systems & Signals Quizzes for GATE EC & GATE EE 5 / 6
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