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I am working on the following problem:

enter image description here

I have a solution that I am pretty confident is correct as have checked the output in python. However, it is long and the entire time I was answering the question I felt like I am missing an obvious, easier approach. Is there a better way to do this?

Solution:

The approach taken is to split the complex signal into cosine and sine components, and evaluate the DFT at $-w_0$ and $+w_0$. The cosine component can be represented as a vector along the real axis with magnitude $AN\frac{1}{2}$

Discretising the signal with $w_0 = 2 \pi f_0$ and $t = nt_s$:

$$ X(w_0) = \sum_{n=0}^{N-1}\cos(w_0 n t_s / N) e^{-iw_0 n t_s / N} = AN\frac{1}{2}e^{0} $$

For the sine component:

$$ \begin{gather*} m_{sin}(nt_s) = - iA(1+\epsilon)\sin(\omega_0 n t_s+\alpha) \\ \\ = -iA(1 + \epsilon) \dfrac{e^{i(\omega_0 n t_s + i\alpha)} - e^{-i(\omega_0 n t_s + i\alpha)}}{2i} \\ \\ = -\frac{1}{2}A(1 + \epsilon) \left[ e^{i\omega_0 n t_s}e^{ i\alpha} - e^{-i\omega_0 n t_s}e^{ -i\alpha} \right]\\ \\ \mathrm{taking \ the\ DFT} \\ \\ X(w_0) = -\sum_{n=0}^{N-1} \frac{1}{2}A(1 + \epsilon)e^{i\omega_0 n t_s / N}e^{ i\alpha} \cdot e^{-iw_0 n t_s / N} + \sum_{n=0}^{N-1} \frac{1}{2}A(1 + \epsilon)e^{-i\omega_0 n t_s / N}e^{ -i\alpha} \cdot e^{-iw_0 n t_s / N} \\ \\ = -\sum_{n=0}^{N-1} \frac{1}{2}A(1 + \epsilon)e^{ i\alpha} + \sum_{n=0}^{N-1} \frac{1}{2}A(1 + \epsilon)e^{-i2\omega_0 n t_s / N}e^{ -i\alpha} \\ \\ = -N\frac{1}{2}A(1 + \epsilon) e^{ia} + \frac{1}{2} A (1 + \epsilon) e^{-ia} \sum_{n=0}^{N-1} e^{-i2\omega_0 n t_s / N} \\ \\ \mathrm{The \ second \ term \ integrates \ to \ zero} \\ = - \left(N\frac{1}{2}Ae^{ia} + N\frac{1}{2}A \epsilon e^{ia} \right) \\ \\ X(w_0) = -N\frac{1}{2}{e^{ia}(A + A\epsilon}) \\ \\ \\ \mathrm{and \ for } -w_0: \\ \\ X(-w_0) = N\frac{1}{2}A(1+\epsilon)e^{-ia} \end{gather*} $$

The $-ia$ exponent is a rotation of the complex phasor in the negative direction, and will not affect the magnitude of the resulting addition with the cosine vector.

Next we can add these complex vector outputs of the DFT and to determine the magnitude, then take the ratio to determine the error $\dfrac{|X(w_0)|}{|X(-w_0)|}$ as required in the question.

$$ \begin{gather*} |X(w_0)| = \sqrt{ (AN\frac{1}{2} - (A+A\epsilon)N\frac{1}{2}\cos(\alpha))^2 + ((A+A\epsilon)N\frac{1}{2}\sin(a))^2} \\ \\ |X(-w_0)| = \sqrt{ (AN\frac{1}{2} + (A+A\epsilon)N\frac{1}{2}\cos(\alpha))^2 + ((A+A\epsilon)N\frac{1}{2}\sin(a))^2} \end{gather*} $$

If we take the ratio $\dfrac{|X(w_0)|}{|X(-w_0)|}$ to find the error and do some cancelling. The final expression I got is:

$$ \begin{gather*} \mathrm{error} = \sqrt{\dfrac{2 - 2\cos(\alpha)(1+\epsilon) + 2\epsilon + \epsilon^2}{2 + 2\cos(\alpha)(1+\epsilon) + 2\epsilon + \epsilon^2}} \end{gather*} $$

so choose $\alpha$ and $\epsilon$ such that error $< 0.001$.

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Your result is correct. A simpler approach would be to follow the hint and write the non-ideal mixing signal as

$$\begin{align}m_{imp}(t)&=A\cos(\omega_0t)-jA(1+\epsilon)\sin(\omega_0t+\alpha)\\&=\frac{A}{2}\left[e^{j\omega_0t}+e^{-j\omega_0t}-(1+\epsilon)\left(e^{j\omega_0t}e^{j\alpha}-e^{-j\omega_0t}e^{-j\alpha}\right)\right]\end{align}$$

Now you can directly see the complex scale factors of the terms at positive and negative frequencies. The quotient of their magnitudes is

$$\frac{\left|1-(1+\epsilon)e^{j\alpha}\right|}{\left|1+(1+\epsilon)e^{-j\alpha}\right|}$$

which equals the expression you arrived at.

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  • $\begingroup$ Thanks! I see my problem was that when trying to follow the hint, I was trying to force both terms into a single complex exponential $Ae^{i\theta}$, which is obviously impossible with the error. $\endgroup$
    – Joseph
    Feb 13, 2023 at 11:33

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