Recent content by noname1

i looked at that however i really don't understand how to do it by sections, in dimensions i tried using a vector with start:end but it always gives me the error above, could you explain how it works?

not sure how to process that in matlab, i know how many samples and spacing per key how can i process just certain samples with fft? i tried fft(singal,[],dim); where dimension is a vector [start:end] however i get an error "Dimension argument must be a positive integer scalar within indexing...

thank you for your reply, i understand how dmtf works however the results i am getting are not making sense to me, for example i am getting this, not sure how to view the higher and lower frequencies

i am trying to figure out digits in a dtmf signal using MATLAB with the fft command however i not sure how to figure out what digit is being pressed. I am doing this stem(abs(signal)) and get a plot with several digits pressed but not sure how to figure out which are pressed, any help would be...

convolve the following h[n] = δ[n-a] + δ[n-b] x[n] = δ[n-c] + δ[n-d] i understand that the convolution is y[n] = h[n]*x[n] and i know how to do it with number instead of letters however not quite sure how it would work with the letters, its not possible to view when a & b will overlap c & d...

nevermind i found my issue forgot to multiply the amplitude

2sin(wot+45)+cis(wot) to Acos(wot) i convert it 2e^-j45 = √2/2 - j√2/2 1e^j0 = 1 - j0 adding these up i get (√2/2 + 1) -j(√2/2) 1.707 - j.707 magnitude = √1.707²+.707² = 1.85 angle = tan^-1(.707/1.707) = 22.5 so i get 1.85cos(wot +22.5) while the book has the answer...

nevermind i found the issue but thanks

yes that is correct it was a typo but the rest looks correct right? than adding these up 5+5*√(3)/2-(5/2) = 6.83 j0+j5/2-j5√(3)/2 = -j1.83 than magnitude would be √6.83²+1.83² = 7.07 and tan^-1 = (-1.83/6.83) = -14.999° = -.08pi

simplfy 5cos(wot+90°)+5cos(wot-30°)+5cos(wot-120°) to form Acos(cos+Ø) i did it this way 5cos(wt) = 5e^j0 = 5+J0 5cos(wot-30°) = 5e^-j30° = 5*(√3/2)+J5/2 5cos(wot-120°) = 5e^-j120° = -5/2*(√3/2)+J5*(√3/2) i have only done partially the problem just missing to add them up however when i look up...

This is not really a homework question just a review but i wanted someone to explain to me why the forces of the moments negative on Fb, Fc and Fd thanks in advanced

this part (5<0° + 14.14<-45°) / 2.24<116.6° = I2 i have my ti-89 calculating it but i just want to know if it is correct if i can calculate polar form like a normal equation example 3x -5 = 10 3x = 15 x=5

this is what i did, i put i2 term on left side and than the rest on the rhs and since i want I2 i divided both equations by 2.24<116.6° 5<0° = -14.14<-45° + I2 * 2.24<116.6° 5<0° + 14.14<-45° = I2 * 2.24<116.6° (5<0° + 14.14<-45°) / 2.24<116.6° = I2 I2 = 8<150.29°

what do you mean by lhs and rhs? i am kind of lost

i2 is being multiplied, its a current in the second mesh but i just wanted to be sure if it is that way its calculated