2
\$\begingroup\$

I have a basic question about down-conversion of RF-signals.

The ideal scenario is described as follows.

Given a RF-signal (coming from an antenna), i.e., a real-valued function x(t), and given a frequency f (say f=100Mhz), one wants to down-convert the band [f-B,f+B] to [-B,+B], say for B=100Khz.

The basic idea is:

  1. to consider x(t) as a complex signal
  2. multiply x(t) by the complex sinusoidal with negative frequency -f, i.e., by \$\cos(-ft) + j \sin(-ft) \$.
  3. This is done in practice by creating two signals \$I(t) = x(t)\cdot cos(ft)\$ and \$Q(t)= x(t)\cdot cos(ft + \dfrac{\pi}{2}) \$
  4. Low-pass-filter I(t) and Q(t) with cutoff frequency B.

The resulting complex signal (i.e., the two signals I(t) and Q(t) ) can then be sampled (by Nyquist, at least at 2*(B+B)=400k samples/sec) with some ADC to do some DSP.

The hardware necessary for doing this appears to be:

  1. Oscillator with frequency f, producing the function cos(f t).
  2. Something to change the phase of the oscillator, to produce \$\cos(f t + \frac{pi}{2}) \$
  3. Two Analog Multiplication units,
  4. One Low-Pass filter with cutoff frequency B.

Question 1: Assuming that the oscillator (perhaps programmable) is given, what kind of hardware would you suggest for (2) and (3) ?

Question 2: Does this setup have significant shortcomings (beside cost of components?)

Since precise multipliers working with high frequencies are expensive, I've read around that one often prefers to multiply x(t) with a complex square wave with frequency f

More precisely,

  1. \$ I(t) = x(t) \cdot Square( f t) \$
  2. \$ Q(t) = x(t) \cdot Square (f t + pi/2)\$

The point is that multiplication by a square wave is just switching, which is probably less expensive to implement!

However the square wave has infinitely many odd harmonics! And therefore it seems to me that the band [-B,+B] of the resulting complex signal:

I(t) + j Q(t)

really is a superposition of the original bands [nf-B, nf+B] of x(t), for all positive odd numbers n, while we wish it to be equal to [f-B,f+B] only!

Question 3: is this observation correct?

To solve the problem, it appears to me that one would have, at the very beginning, to LOW-PASS the signal x(t) with cutoff frequency f+B.

Question 4: Having an oscillator with programmable frequency f is realistic. But how do we implement a LOW-PASS filter with variable (f+B) cut-off frequency [the variable is f]?

In the schematics I've found online (e.g. Wikipedia) there is not mention of this variable LOW-PASS filter.

\$\endgroup\$
  • 1
    \$\begingroup\$ Have you looked at heterodyne and superheterodyne recievers? Note that wikipedia article talks about doing the processing in pure digital, applying a fixed low-pass filter; this is sufficient provided the downconverted frequency is sufficiently far from the carrier. \$\endgroup\$ – pjc50 May 6 '14 at 21:24
  • 2
    \$\begingroup\$ Welcome to EE.SE. You might get better answers if you split your question into smaller chunks. \$\endgroup\$ – David May 6 '14 at 22:11
  • 1
    \$\begingroup\$ For your question 3, this question might be interesting. \$\endgroup\$ – AndrejaKo May 6 '14 at 22:35
  • \$\begingroup\$ You need an IQ modulator and a Direct Digital Synth (DDS) \$\endgroup\$ – hassan789 May 7 '14 at 0:58
2
\$\begingroup\$

Question 1: Assuming that the oscillator (perhaps programmable) is given, what kind of hardware would you suggest for (2) and (3) ?

For 2: All-pass (phase-shifting) filter, a.k.a. Hilbert transformer

For 3: Balanced modulator

Question 2: Does this setup have significant shortcomings (beside cost of components?)

Yes. While direct conversion to complex baseband is regularly done in the digital domain, it can be quite finicky to get the same concept working well (and reliably) in the analog domain. It's nearly always easier to do the detection at an intermediate RF frequency.

Question 3: is this observation [about square-wave local oscillators] correct?

Yes, except that you want a bandbass filter that's centered on the carrier frequency, not a lowpass filter. Sometimes this filter has a bandwidth of approximately 2B, in which case it needs to be tuned along with the local oscillator. This sort of setup is called a preselector, and is commonly used in superheterodyne receivers, such as those used for the AM and FM broadcast bands.

But sometimes, a fixed bandpass filter that covers the entire band of interest is used instead. This works as long as none of the unwanted "image" bands ever falls into the band of interest. This is more common in 2-way VHF communications systems, such as those used for air traffic control and public service (police, fire).

\$\endgroup\$
  • \$\begingroup\$ Thank you very much! Very useful. I understand from your Answer 1 that you also suggest to use a modulator (i.e., multiplication by a square) rather than a multiplication! THANKS! \$\endgroup\$ – IamMeeoh May 7 '14 at 9:38
0
\$\begingroup\$

Question 1: Assuming that the oscillator (perhaps programmable) is given, what kind of hardware would you suggest for (2) and (3) ?

I would suggest a dual DDS running out of phase I+Q duties (analog devices have dual synched dds), you have to convert DDS stepped sine to square wave using a hard driven op amp or hard limiter. I would use fast acting rf switches (non terminated) for the mixers so you really dont need a lot of rf gain.

Question 2: Does this setup have significant shortcomings (beside cost of components?)

In this arrangement you are not using a non linear device to drive multiplication but using the phase flips to drive multiplication, thus less IMD products. The only drawback with zero IF and low IF receiver designs are microphonics with too much LO energy leakage.

\$\endgroup\$

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.