2. Use the result to compute the Laplace transform of the half-wave rectified sine signal given by, sin wt 2nn

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Chapter3: Functions
Section3.3: More On Functions; Piecewise-defined Functions
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III) Let function f be periodic with period T > 0, that is, f(t) = f(t+T), for -o <t<
оо. Define
•T
F:(+) = [ c*;
e-st f(t)dt.
1. Show that the Laplace transform of f is F(s) = Fi(s).
Hint: Split the domain [0, ∞0] as [0, T]U [T, ∞0]; for the second integral, do a change
of variable to make the integral look like the standard Laplace integral.
2. Use the result to compute the Laplace transform of the half-wave rectified sine signal
given by,
(2n+1)т
{
sin wt
2nn
<t<
f(t) =
(2n+1)T
(2n+2)T
<t<
n = 0, 1, 2, ...
You may se:
eat
| eat sin(wt)dt
:(-w cos(wt) + a sin(wt).
a² + w?
Transcribed Image Text:III) Let function f be periodic with period T > 0, that is, f(t) = f(t+T), for -o <t< оо. Define •T F:(+) = [ c*; e-st f(t)dt. 1. Show that the Laplace transform of f is F(s) = Fi(s). Hint: Split the domain [0, ∞0] as [0, T]U [T, ∞0]; for the second integral, do a change of variable to make the integral look like the standard Laplace integral. 2. Use the result to compute the Laplace transform of the half-wave rectified sine signal given by, (2n+1)т { sin wt 2nn <t< f(t) = (2n+1)T (2n+2)T <t< n = 0, 1, 2, ... You may se: eat | eat sin(wt)dt :(-w cos(wt) + a sin(wt). a² + w?
1) Given that f is periodic with period T > 0 that is,f (t) = f (t + T)
T
Define F1 (s) = 6 e-stf (t)dt
The objective is to show that F (s) =
F1 (s).
1-e-sT
The formula for Laplace transform is,
F (s)= e-"f (1)dt
00
=6' e-f (1)dt + f e-f (1)dt
=F1 (s) + e-if (f)dt
Step 2
Now take V = t – T so, dV = dt.
Now substitute in the F (s) = F 1 (s) + Jr e ~" f (t) d t,
00
F (s) = F1 (s) + e-s(V+T)f (V + T)dV
Since it is known thatf is periodic so f (V + T) = f (V).
F (s) = F1 (s) + e-T L® eVƒ (V)dV
F (s) = F1 (s) + e¬sTF (s)
(1- e-s")F (s) = F1 (s)
%3D
F (s) = (1)F1 (s)
1-e-sT
Transcribed Image Text:1) Given that f is periodic with period T > 0 that is,f (t) = f (t + T) T Define F1 (s) = 6 e-stf (t)dt The objective is to show that F (s) = F1 (s). 1-e-sT The formula for Laplace transform is, F (s)= e-"f (1)dt 00 =6' e-f (1)dt + f e-f (1)dt =F1 (s) + e-if (f)dt Step 2 Now take V = t – T so, dV = dt. Now substitute in the F (s) = F 1 (s) + Jr e ~" f (t) d t, 00 F (s) = F1 (s) + e-s(V+T)f (V + T)dV Since it is known thatf is periodic so f (V + T) = f (V). F (s) = F1 (s) + e-T L® eVƒ (V)dV F (s) = F1 (s) + e¬sTF (s) (1- e-s")F (s) = F1 (s) %3D F (s) = (1)F1 (s) 1-e-sT
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