Reply by Randy Yates March 20, 20112011-03-20
On 03/20/2011 10:14 AM, colsandurz45 wrote:
> Hello, > > I'm not sure I fully grasp quadrature sampling. My idea of what quadrature > sampling is (from a communications perspective) this: > > You start with two ADCs, one to sample in-phase (I) and one to sample > quadrature (Q). The two ADCs sample 90 degrees out of phase. > > So if I have a signal x(t) = sin(2*pi*f*t) + cos(2*pi*f*t) and I sample it > with my two ADCs that are 90 degrees out of phase at a a rate f. Can I > fully recovery x(t)? I can't think of a reason why not.
If you have a real signal, you can either use one ADC sampling at twice the highest frequency or two ADCs sampling at the highest frequency if you do some manipulation of the sampling and/or input signal. The scenario you described falls under the former.
> Also, if my understanding of quadrature sampling is correct, is there a > difference between quadrature sampling at a rate fs and normal sampling at > a rate 2*fs (assuming the sampled signal is complex)?
"Quadrature sampling" is the process of sampling a complex signal in rectangular form. That is, if z(t) = x(t) + i*y(t), then z[n] = x[n] + i*y[n]. I[n] is x[n] and Q[n] is y[n]. There are a myriad of ways to derive a complex signal, e.g., a complex mix to an IF, a complex mix to baseband, sampling the same signal at different sampling phases, etc. I have a suprise for you (at least it was a surprise for me when I realized it): sampling ALWAYS gives you a bandwidth of -Fs/2 to +Fs/2, whether real or complex. However, real sampling forces half of that bandwidth to be redundant, or in other words, to carry no extra information. Complex sampling allows the entire bandwidth to carry information. So, everything you always wanted to know about "quadrature sampling" is available if you just look at it in terms of real/complex numbers and standard linear system theory (e.g., Fourier transforms, etc.). -- Randy Yates Digital Signal Labs 919-577-9882 http://www.digitalsignallabs.com yates@digitalsignallabs.com
Reply by dbd March 20, 20112011-03-20
On Mar 20, 7:14&#4294967295;am, "colsandurz45" <dwwkelly@n_o_s_p_a_m.gmail.com>
wrote:
> Hello, > > I'm not sure I fully grasp quadrature sampling. &#4294967295;My idea of what quadrature > sampling is (from a communications perspective) this: > > You start with two ADCs, one to sample in-phase (I) and one to sample > quadrature (Q). &#4294967295;The two ADCs sample 90 degrees out of phase. > > So if I have a signal x(t) = sin(2*pi*f*t) + cos(2*pi*f*t) and I sample it > with my two ADCs that are 90 degrees out of phase at a a rate f. &#4294967295;Can I > fully recovery x(t)? &#4294967295;I can't think of a reason why not. > > Also, if my understanding of quadrature sampling is correct, is there a > difference between quadrature sampling at a rate fs and normal sampling at > a rate 2*fs (assuming the sampled signal is complex)?
"Quadrature sampling" is a term that can refer to a number of concepts. Try googling, the first 50 hits give discussions of a number of these concepts. Dale B. Dalrymple
Reply by colsandurz45 March 20, 20112011-03-20
Hello,

I'm not sure I fully grasp quadrature sampling.  My idea of what quadrature
sampling is (from a communications perspective) this:

You start with two ADCs, one to sample in-phase (I) and one to sample
quadrature (Q).  The two ADCs sample 90 degrees out of phase.

So if I have a signal x(t) = sin(2*pi*f*t) + cos(2*pi*f*t) and I sample it
with my two ADCs that are 90 degrees out of phase at a a rate f.  Can I
fully recovery x(t)?  I can't think of a reason why not.

Also, if my understanding of quadrature sampling is correct, is there a
difference between quadrature sampling at a rate fs and normal sampling at
a rate 2*fs (assuming the sampled signal is complex)?