Re: Signal coming from the USRP to the computer

I think a little more detailed precise answer to John’s question might
help:

John A. wrote:

each complex sample that enters the
USB bus is the following,

x[i] = (inphase_component) + j (quadrature_component), and
x[i] = m(t)cos( 2piFREQ_OFFSETt + PHI ) + jm(t)sin(
2
piFREQ_OFFSETt +
PHI ), where m(t), is the actual message signal, FREQ_OFFSET is the
frequency offset, and PHI is the phase.

Is that correct?

I think you’re confusing the baseband and passband signals a little, and
the equations aren’t quite right.

The complex-baseband signal (your message) is the data that is
transferred
across the USB channel.
x[i] = (in-phase) + j*(quadrature)
= (x_i) + j*(x_q)

These are samples of your message signal, after modulation (mapping to a
complex QAM-constellation, for example), coding, pulse-shaping, etc.

The signal is up/down converted on the USRP device such that the
transmitted RF signal is

r(t) = x_icos(2pif_c) - (x_q)sin(2pif_c)

(where f_c is your RF carrier frequency, and I’m ignoring phase offsets
and noise)

Notice the subtraction there (which comes from the trig identities) and
that all the terms are real (it’s a real passband signal).

Hope that helps a little.

Patrick Sisterhen
National Instruments

Thanks Patrick. I was concerned with the received signal path. Suppose,
I
have the receiver tuned to, let’s say, GPS signal. What will the
received
signal look like. Considering the GPS message signal is m(t), then what
would equation would best describe the received signal.

If ‘f_c’ is the carrier frequency then the signal coming over the USB
bus on
to the computer for baseband processing will be,
inphase(t) = m(t) cos(phi)
quadrature(t) = m(t)sin(phi)

where, ‘phi’ is the instantaneous offset. Remember, phi here is a broad
term
which includes all kinds of offsets(frequency, phase etc).

On Tue, May 31, 2011 at 11:47 AM, Patrick Sisterhen <

John,

Typo in my equations, should have been:

y_q = (x_i * sin(phi)) - (x_q * cos(phi))

Patrick Sisterhen
National Instruments

From: John A. [email protected]
To: Patrick Sisterhen [email protected]
Cc: [email protected]
Date: 05/31/2011 02:08 PM
Subject: Re: [Discuss-gnuradio] Signal coming from the USRP to
the
computer

Thanks Patrick. I was concerned with the received signal path. Suppose,
I
have the receiver tuned to, let’s say, GPS signal. What will the
received
signal look like. Considering the GPS message signal is m(t), then what
would equation would best describe the received signal.

If ‘f_c’ is the carrier frequency then the signal coming over the USB
bus
on to the computer for baseband processing will be,
inphase(t) = m(t) cos(phi)
quadrature(t) = m(t)sin(phi)

where, ‘phi’ is the instantaneous offset. Remember, phi here is a broad
term which includes all kinds of offsets(frequency, phase etc).

On Tue, May 31, 2011 at 11:47 AM, Patrick Sisterhen <
[email protected]> wrote:
I think a little more detailed precise answer to John’s question might
help:

John A. wrote:

each complex sample that enters the
USB bus is the following,

x[i] = (inphase_component) + j (quadrature_component), and
x[i] = m(t)cos( 2piFREQ_OFFSETt + PHI ) + jm(t)sin(
2
piFREQ_OFFSETt +
PHI ), where m(t), is the actual message signal, FREQ_OFFSET is the
frequency offset, and PHI is the phase.

Is that correct?

I think you’re confusing the baseband and passband signals a little, and
the equations aren’t quite right.

The complex-baseband signal (your message) is the data that is
transferred
across the USB channel.
x[i] = (in-phase) + j*(quadrature)
= (x_i) + j*(x_q)

These are samples of your message signal, after modulation (mapping to a
complex QAM-constellation, for example), coding, pulse-shaping, etc.

The signal is up/down converted on the USRP device such that the
transmitted RF signal is

r(t) = x_icos(2pif_c) - (x_q)sin(2pif_c)

(where f_c is your RF carrier frequency, and I’m ignoring phase offsets
and noise)

Notice the subtraction there (which comes from the trig identities) and
that all the terms are real (it’s a real passband signal).

Hope that helps a little.

Patrick Sisterhen
National Instruments

John,

A phase difference (phi) between the frequency of the transmit carrier
(f_c) and the receiver local oscillator (f_r) will be exhibited as a
rotation of your received symbols in the complex plain.

I think that’s what you mean to imply in your equations, but to get a
little more precise:

Let x_i and x_q be the in-phase and quadrature components of your
baseband
message at the transmitter.
Let y_i and y_q be the same for the baseband message at the receiver
(after downconversion with an oscillator with a phase offset).

y_i = (x_i * cos(phi)) + (x_q * sin(phi))
y_q = (x_i * cos(phi)) - (x_q * cos(phi))

If phi is 0, you recover the original in-phase and quadrature
components.
Otherwise, it works like a rotation by phi.

If the receiver local oscillator has a frequency offset from the
transmitter oscillator (f_c != f_r), the received symbols will
continuously rotate over time.

(I may have reversed the sign in those equations… it depends on the
implementation of the downconverter, but you can detect and correct for
it
in the same way.)

Patrick Sisterhen
National Instruments

From: John A. [email protected]
To: Patrick Sisterhen [email protected]
Cc: [email protected]
Date: 05/31/2011 02:08 PM
Subject: Re: [Discuss-gnuradio] Signal coming from the USRP to
the
computer

Thanks Patrick. I was concerned with the received signal path. Suppose,
I
have the receiver tuned to, let’s say, GPS signal. What will the
received
signal look like. Considering the GPS message signal is m(t), then what
would equation would best describe the received signal.

If ‘f_c’ is the carrier frequency then the signal coming over the USB
bus
on to the computer for baseband processing will be,
inphase(t) = m(t) cos(phi)
quadrature(t) = m(t)sin(phi)

where, ‘phi’ is the instantaneous offset. Remember, phi here is a broad
term which includes all kinds of offsets(frequency, phase etc).

On Tue, May 31, 2011 at 11:47 AM, Patrick Sisterhen <
[email protected]> wrote:
I think a little more detailed precise answer to John’s question might
help:

John A. wrote:

each complex sample that enters the
USB bus is the following,

x[i] = (inphase_component) + j (quadrature_component), and
x[i] = m(t)cos( 2piFREQ_OFFSETt + PHI ) + jm(t)sin(
2
piFREQ_OFFSETt +
PHI ), where m(t), is the actual message signal, FREQ_OFFSET is the
frequency offset, and PHI is the phase.

Is that correct?

I think you’re confusing the baseband and passband signals a little, and
the equations aren’t quite right.

The complex-baseband signal (your message) is the data that is
transferred
across the USB channel.
x[i] = (in-phase) + j*(quadrature)
= (x_i) + j*(x_q)

These are samples of your message signal, after modulation (mapping to a
complex QAM-constellation, for example), coding, pulse-shaping, etc.

The signal is up/down converted on the USRP device such that the
transmitted RF signal is

r(t) = x_icos(2pif_c) - (x_q)sin(2pif_c)

(where f_c is your RF carrier frequency, and I’m ignoring phase offsets
and noise)

Notice the subtraction there (which comes from the trig identities) and
that all the terms are real (it’s a real passband signal).

Hope that helps a little.

Patrick Sisterhen
National Instruments

Yes Patrick, I agree with your explanation. That was succinct and easy
to
understand. Thanks.

On Tue, May 31, 2011 at 4:55 PM, Patrick Sisterhen
<[email protected]

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