Recent content by skan

plotting phase and magnitude - urgent hi, Using the Fourier transform integrals i find the Fourier transform of a function to be x(w) = j (5To/2) { sinc [5To/2 (W+Wo)] - sinc[5To/2 (W-Wo)] } Can someone pls tell me how to find the phase and the magnitude from this equation and how to...

I mean I read in my notes that the integral of a impulse function is a unit step function

Sorry for the double post. 1.Thanks but I take the FT of c(t) first and then I shift all the impulse functions by wa. can someone please explain why as i don't understand why this is done this way. 2.Also I thought that the Integral of an impulse function is 1, but here its says that the...

hi , the question is to find the Fourier transform of the following eqn without using integration d(t)= [c(-1.5t)]exp(jwat) where c(t)= Ao + E(Sumation)An*cos(nwot) + Bn sin(nwot) [fourier series formula] I knwo to find the FT of the above Fourier series. But why is exp(jwat)...

thanks a lot

hi, How do u find the e^6 or the exponential of 6 or -6 or any number in a scientific calculator? Thanks skan

yeah I am ! they are filters.

I am having troubel understanding slopes. I know the slope formulas but when given a triangle function, say on the x-axis one end of the base of the triangle is at -2 and the other end at 2 and on the y-axis they converge at 2. and given a rect function, and when asked to find the area of...

1.g(t) = f(t) + f(-t) 2. g(t)= f(t/2) can someone pls explain why these functions are time variant and non causal

1.g(t) = f(t) + f(-t) 2. g(t)= f(t/2) can someone pls explain why these functions are time variant and non causal

thanks a lot

f(t) = 1 if -pi/2 <=t <=0 -1 if 0<=t<= pi/2 0 elsewhere how does integral of f(t)*cost dt become and odd function with the integral limit from - pi/2 to pi/2 ? thanks a lot

Thanks a lot for the answer and the great link!

I have basic trignometry doubt how to u get 5cos(2t-53.13degrees) from 3cos2t + 4 sin2t. Can someone suggest a site from where I can lean basic trinometry thanks. skan