I have an exercise from a textbook about smart antennas. I think I have the right approach to solve this kind of problem. However, I'm not quite sure if I'm handling it correctly due to the additional phase delay of one radiator.

The exercise looks like this:

Consider two isotropic radiators located at a distance of d=0.55λ. Both radiators are driven by a sinusoidal carrier signal with equal amplitude. The phase of the signal feeding radiator 1 lags behind the signal of radiator 2 by 60°. Calculate the angle α, at which there is a zero of the radiation.

So the setup looks like this: enter image description here

My approach would be adding two complex signals togehter like this \begin{equation} \underline{g}(\phi)=\sum_{n=0}^{N-1} \underline{w}_{n} \cdot g_{n}(\phi) \cdot \exp (-j \cdot k \cdot n \cdot d \cdot \sin \phi) \end{equation}

As far as I understand \$g_n(\phi)=1\$ because the antennas are isotropic and \$k = \frac{2\pi}{\lambda}\$ because it is the propagation constant.

\$\omega_n\$ has to be the sinusoidal which should look like \$A\cdot cos(2\pi f t+\phi)\$. Here I am not sure. This is the correct way of adding the phase right? And do I assume correctly that I cannot get a numerical result because there was never mentioned which frequency the signals have?

Since the formula above looks like a Fourier transformation I assume one can make use of it. I don't see how. Transforming two sinusoidal to 4 peaks, adding them and transforming them back doesn't help much.

In the end I get something like this: $$g(\phi)=cos(2\pi ft)exp(0)+cos(2\pi ft-60°)exp(-j2\pi0.55 sin(\phi))$$ which I then would set equal to zero. Whilst writing this I just noticed that I have introduced a variable t. No clue how to handle this. All my exercises before did not depend on the frequency nor on time. Why would the pattern in two dimensions not depend on the signal and the time? In all the baby examples we did before there was an easier formula if all antennas were equally spaced: \begin{equation} \Delta \varphi=-k \cdot n \cdot d \cdot \sin \phi \end{equation}

First I thought this might be applicable since I could convert the phase delay in an additional distance. But I guess this is only true in x direction. I just got more confused writing this all. Guess I haven't understood it at all.


you can work in degrees or radians using \$\lambda\$ so frequency is normalized and not required.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.