Not really sure what you are trying to show here. Wouldnt a constant phase angle simplify things a bit?Mr Al,
You are correct about the phase factor, but look more carefully at what he wrote. There are two impulse functions, so the changing phase factor with frequency only matters at the frequencies (there are two of them) where the impulse functions are. You will notice that you get +θ or -θ at each of those impulse functions.
Actually it appears that he did not do this yet, and is multiplying in frequency, or intends to? Maybe not as i do see the little asterisk there with the note. Maybe i read that wrong.Also, you are correct about multiplication in the time domain is convolution in the frequency domain, but it's not clear why you point this out. This appears to be what the OP did.
Not really sure what you are trying to show here. Wouldnt a constant phase angle simplify things a bit?
The * is a standard symbol for convolution. This is a it confusing because * is also used for multipication in computer programing languages. But, the math convention preceeded the programing language convention, I think.Actually it appears that he did not do this yet, and is multiplying in frequency, or intends to? Maybe not as i do see the little asterisk there with the note. Maybe i read that wrong.
So you are saying he actually intends to do a convolution in frequency rather than A(f) times C(f) ?
The imaginary part is important, but we always get complex conjugate pairs when the time domain signal is real.Also, i dont remember what we do with the imaginary part, if any, that could come up in the transform such as F(-sin(wt)).
I didn't check the transform of c(t) carefully, but it looks right, and certainly the structure looks correct. I can check it more carefully later, but the structure is what matters for the method of calculating the answer. You have a sum of two impulse functions in the frequency domain. I think the frequency domain typically is in terms of omega (2pi f) rather than f, but maybe that is just the way you prefer to do it, in terms of f.
Next you need to find the transform of the a(t), which is a square wave. This will be an infinite sum of impulse functions in the frequency domain. Once, you have both A(w) and C(w) correct, you will be ready to find the convolution of the two.
Convolution of impulse functions are relatively easy because the convolution integral is sliding one function past the other, and only when the impulses overlap is there a non-zero value. Since one function only has a sum of 2 impulses, break the problem into a sum of separate convolution integrals, each with one impulse function only from C(w). Then you will see that each of the two integrals is easy to calculate as one impulse slides through the infinite sum of impulses in A(w). Then add the result of the two integrals together to get A(w) * C(w). Transforming back to the time domain will give the sum of two sine waves, and I think you will basically discover a trig identity in the time domain. This will allow you to check your answer. Using the trig identity on c(t) a(t) should give you the same answer.
The other way to solve the answer for A(w) * C(w) is to use the trig identity first on a(t) c(t) and then transform that back to the frequency domain.
Each impulse funtion has its own phase factor, or complex coefficient. If phi=0, then the two constants are equal. If phi=90 degrees, then they are equal in magnitude, and oposite in sign. So, there is really no frequency dependence to the phase factor, but the f/Fc basically encodes a plus and minus sign, if we think of sine and cosine only. I agree it would be simpler to put the negative sign with the left side impluse function, in those cases, or whenever possible. But, the notation shown is compact.
The * is a standard symbol for convolution. This is a it confusing because * is also used for multipication in computer programing languages. But, the math convention preceeded the programing language convention, I think.
The imaginary part is important, but we always get complex conjugate pairs when the time domain signal is real.
The imaginary part is important, but we always get complex conjugate pairs when the time domain signal is real.
Yes, but that's not quite right. The transform for sin(wt) has a negative sign on one of the impulses. We can see this using Euler's relations.What i meant was the transform for cos(wt) is:
pi*Impulse(w,w0)
while the transform for sin(wt) is
pi/j*Impulse(w,w0)
One is real while the other is imaginary.
Yes, but that's not quite right. The transform for sin(wt) has a negative sign on one of the impulses. We can see this using Euler's relations.
2 cos(x) = exp(jx) + exp(-jx)
2j sin(x) = exp(jx) - exp(-jx)
So, what I was saying is that the coefficients of the impulses are always complex conjugates of one another. One is exp(-j phi) and the other is exp(+j phi). When phi=0, then the coefficients are equal (to 1), and when phi=pi/2, then the coefficients are +j and -j. For other angles, they are two complex numbers that are complex conjugates of each other.
Getting back to your original point, I think what is bothering you is the "f/Fc" that shows up in the formula. All this does is encode the plus and minus sign so that the coefficients are complex conjugates of one another. It is probably less confusing to just put the exp(j phi) and exp(-j phi) in front of each impulse function, rather than trying to have one overall exp(f phi / Fc) that applies to both impulse functions.
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