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This is really stumping me. I have a signal that is being modulated at frequency $f$, and is sampled every $\Delta t$ seconds for $N$ samples. I'll index these samples with $k$.

The signal is $S(k\Delta t) = -A \cos(2 \pi f k \Delta t)$

The derivative of this is approximately: $S'(k \Delta t) = \frac{S((k+1) \Delta t) - S(k \Delta t)}{\Delta t} = -\frac{A}{\Delta t}(\cos(2 \pi f (k+1) \Delta t)-\cos(2 \pi f k \Delta t))$

By writing these out in complex exponential form, I can rewrite this as:

$S'(k \Delta t) = -\frac{A}{2 \Delta t}(e^{2\pi i f \Delta t}+1)\cos(2 \pi f k \Delta t)$.

However, I know that the derivative of a cosine should be a negative sine! Probably:

$S'(k \Delta t) = 2\pi f \Delta t A \sin(2 \pi f k \Delta t)$.

How on earth do I show that these last two lines are approximately equal? Thank you very much for your help!

Bunji
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  • I think that you made some mistake in going from your first equation for S' to your second equation for S'. In your first equation it's apparent that S' should be a real number for any value of delta(t). But for your second equation for S', the result will in general be some complex number for most values of delta(T), including small values, due to the presence of the exponential term. – Barney Cowell May 11 '17 at 18:09

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Just copy the proof of the derivative for cosine, and replace the limits with approximations. I assume $f$ and $A$ are constant frequency/amplitude.

$$-\frac{A}{\Delta t}(\cos(2 \pi f (k+1) \Delta t)-\cos(2 \pi f k \Delta t))$$

Since $\cos(\alpha+\beta) = \cos \alpha \cos \beta – \sin \alpha \sin \beta$, that becomes ($\alpha = 2 \pi f k \Delta t, \beta = 2 \pi f \Delta t$):

$$-\frac{A}{\Delta t}(\cos(2 \pi f k \Delta t)\cos(2 \pi f \Delta t) - \sin(2 \pi f k \Delta t)\sin(2 \pi f \Delta t)-\cos(2 \pi f k \Delta t)) = -A\left(\cos(2 \pi f k \Delta t)\frac{\cos(2 \pi f \Delta t) - 1}{\Delta t} - \sin(2 \pi f k \Delta t)\frac{\sin(2 \pi f \Delta t)}{\Delta t}\right)$$

As of yet, we've only used algebra on the discrete derivative. At this point, we use the following approximations, based on well-known limits that can be found in the same trig identity wiki article:

$$\frac{\cos(2 \pi f \Delta t) - 1}{2 \pi f\Delta t} \approx 0 \qquad \frac{\sin(2 \pi f \Delta t)}{2 \pi f\Delta t} \approx 1$$

Therefore, the above equation can be approximated as:

$$A2 \pi f \sin(2 \pi f k \Delta t)$$

As required.

Larry B.
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