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# Need for I-Q modulator and demodulator

by on April 5, 2008

Some time back, a friend mentioned that he is yet to understand the need for having both in-phase (I) and quadrature-phase (Q) signals in typical wireless systems.

In this post, the attempt is to bring out the motivation for having I-Q modulation and present the block diagram of a simple I-Q modulator (and demodulator).

I am using the text provided in Sec 5.2.1 of [DIG-COMM-BARRY-LEE-MESSERSCHMITT] as reference.

## Baseband PAM transmission

Consider a simple baseband transmission where the information is sent by modulating a pulse. This can be represented as

$s(t) = \sum_{m=-\infty}^{\infty}a_mg(t-mT)$, where

$T$is the symbol period,

$a_m$ is the symbol to transmit,

$g(t)$ is the transmit filter,

$m$ is thesymbol index and

$s(t)$is the output waveform.

Pictorially the same can be represented as,

Figure: Baseband PAM transmission (Refer: Figure 5.1 [DIG-COMM-BARRY-LEE-MESSERSCHMITT]]

To transmit $s(t)$ undistorted through the channel, the minimum bandwidth required is half the symbol rate $frac{1}{2T}$ (per the Nyquist criterion).

Figure: Minimum bandwidth required for transmitting baseband PAM with symbol rate $frac{1}{T}$

Assuming that there are $N$ alphabets in $a_m$, the spectral efficiency for basedband PAM is,

$Gamma_{BB-PAM}={\frac{\log_2N}{\frac{1}{2T}}}/T=2T\log_2N/T = 2\log_2N$bits/second/Hertz.

BB-PAM : Baseband PAM

## Passband PAM transmission

Now, consider that the baseband PAM signal is upconverted to a carrier frequency $f_c$ by multiplication by a carrier, i.e.

$hat{s}(t) =\sqrt2cos(2\pi f_c t) \sum_{m=-\infty}^{\infty}a_mg(t-mT)$.

The spectrum of the passband PAM is as shown below,

Figure: Spectrum of passband PAM

To avoid transmit $hat{s}(t)$ undistorted through the channel, we require a passband filter having bandwidth $frac{1}{T}$.

Assuming that there are $N$ alphabets in $a_m$, the spectral efficiency for passband PAM is,

$Gamma_{PB-PAM}={\frac{\log_2N}{\frac{1}{T}}}/T=T\log_2N/T = \log_2N$bits/second/Hertz.

PB-PAM : PassBand PAM

Can see that spectral efficiency of passband PAM is half of baseband PAM.

Note:

Passband PAM is composed of real signals and real signals have symmetric spectra. So, half the bandwidth is not carrying any ‘extra’ information.

## Moving to passband QAM

It is known that if a sine wave and a cosine wave is periodic over time $T$, then they are orthogonal i.e.
$\int_T \sin wt \cos wt = 0$.

Given so, a popular way which people came across to improve the efficiency of the passband PAM is to send information on the sine wave also in parallel, i.e.

$\bar{s}(t) =\sqrt2\cos (2\pi f_c t) \sum_{m=-\infty}^{\infty}a^I_mg(t-mT) - \sqrt2\sin (2\pi f_c t) \sum_{m=-\infty}^{\infty}a^Q_mg(t-mT)$.

This forms the passband QAM modulation. This signal requires the same bandwidth as passband PAM, however carries twice the informtion. For notational convineance, the above signal can be represented mathematically as,

$\bar{s}(t)=\sqrt{2}\Re \left[ \tilde{s}(t)e^{j\2\pi f_c t}\right]$, where

$\tilde{s}(t)=\sum_m \tilde{a}_mg(t-mT)$ and

$tilde{a}_m=a^I_m+a^Q_m$.

This forms the I-Q modulator circuit.

Figure: IQ modulator

The IQ demodulator can be visualized as shown in the figure below

Figure: IQ demodulator

On the combined signal, downconvert the cosine (inphase, I) and the sine (quadrature Q) arms, then proceed to do independent demodulation on each arm.

Summarizing,

(a) It is the need for improved spectral efficiency that resulted in I-Q modulation (and de-modulation).

(b) Another approach for improving the spectral efficiency of passband PAM is to filter away half the bandwidth (which is not carrying any ‘extra’ information). This is called Single Sideband Modulation (SSB).

Anyhow, it seems that using I-Q modulation is simpler to implement than circuit for filtering away half the spectrum. So, I-Q modulation stays.

Reference

Thanks,
Krishna

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Maxsteel August 28, 2012 at 8:27 pm

Hi Everyone,

I think I,Q transmission and receive system is just one way of baseband data transmission. Any baseband signal can be broken up into I and Q components. There are other forms of transmitters also known as polar transmitter. But in real circuit implementation polar transmitters are hard to build. Its very tough to change the phase of carrier precisely. But by using I,Q one can adjust the phase just by varying amplitudes of I,Q.

Krishna Sankar August 29, 2012 at 5:15 am

@Maxsteel: Thanks

Thulasi February 7, 2012 at 10:56 pm

written in simple way.. easy to understand:) it helps a lot.. thank you:)

komal November 3, 2010 at 5:25 pm

Hi!. How the source coding can compress the data rate?And how the modulation schemes are able to increase the data rate,though we are representing the symbol with redundant bits?Can u plz put forward the process of digital communication with an example?

Krishna Sankar November 15, 2010 at 2:12 am

@komal:
a) Source coding : If we have redundancy in the information symbol, we can have a better way of sending that information out (rather than sending the raw data). For eg, if we have a sequence [0 0 0 0 0 1 1 1 ], we can alternatively send [ 4 zeros three ones]
b) Modulation : We can pack more bits into a symbol. However, the penalty is that we need higher signal to noise ratio to recover the bits.

anil kumar January 20, 2010 at 1:47 am

Hi evry one in this.Nice to c u all here buddy’s.
Ofcourse I am new to this. I have a small problem with my calculation in OFDM USING QAM AND IQ modulator .
I ll just paste the question here. I have done some calculations, but i was unable to judge those.Please help as early as possible.

We are required to build a SIMULINK model of a 16-QAM OFDM Digital Communications System (OFDM transmitter and OFDM receiver) operating over an AWGN channel.
• The OFDM system has a cyclic prefix of 20% of the (data) symbol duration Ts.
• Use an I-Q modulator at the transmitter and I-Q demodulator at the receiver to upconvert and downconvert, respectively, the analogue signal to a central frequency of fc=2 MHz (so, the passband transmitted signal is centered on 2 MHz).
• The symbol duration is set to Ts =1 msec.
• Chose the number of carriers, N, so as to achieve a bitrate of Rb = 192000 bits / sec (you have to evaluate the correct N).
• Test the system performance for Eb/N0=5, 8, 10, 12 and 14 dB, and report the respective BER measurements.
• Subsequently, use a channel encoder of your choice and repeat the BER measurements for the same values of Eb/N0.

kwame November 21, 2009 at 2:03 pm

Hi krishna,

good work on your posts. I have a small question on the IQ modulator, Since we were previously not utilizing the other half of the spectrum in passband transmission and hence introduced the IQ scheme to fully use the bandwidth, does it mean we are actually sending the same data (data symbol) in both I and Q arm? i.e if its a1, then its a1 carried in both arms?

Krishna Sankar December 6, 2009 at 4:25 pm

@kwame: Well, its not that we were not utilizing the other half of the spectrum. Rather the same information (mirrored in frequency) was present in the other half of the spectrum. In previous case, we were not sending anything on the Q arm.

mhamdi November 5, 2009 at 1:39 pm

can you give me some programm for cognitive radio

Krishna Sankar November 8, 2009 at 8:55 am

@mhamdi: Sorry, I do not have posts on cognitive radio

Umair August 9, 2009 at 5:04 pm

Hi,…can anybody guide me in making an FM transmitter and receiver on Simulink and System Generator and its implementation on Lyrtech SFF SDR DP…. i dont kno how to do it for implementation kindly do help me

Krishna Sankar August 11, 2009 at 4:45 am

@Umair: Sorry, I do not have experience with Simulink. Anyhow, am planning to write a post on FM transmission/reception in the near future.

pradeep February 15, 2009 at 9:07 am

thanks for replying bhayya..ya actually now a days they r using it seems.. actually am doin my final year of engg.. and am doin a project based on this.. i have a thought.. can u inform ur mail id so that i can put it up to u and u can suggest the necessary corrections bhayya..i ll be pleased if u can help me out.. hope am not disturbing u…

pradeep February 4, 2009 at 5:28 pm

hi…. the article on iq modulator was good… i have a small doubt. how can we generate a FM signal usig this iq modulator

Krishna Sankar February 10, 2009 at 7:23 pm

@pradeep: Thanks. I have not studied the FM transmitter well enough to comment. Just a quick thought: do we need I/Q modulator for FM.

MY MAN January 29, 2009 at 2:29 pm

Yes.. the last post very informative. Please incorporate more pictures and explanations than the MATH. Great work though!!

Krishna Sankar January 30, 2009 at 5:42 am

@MY MAN: Thanks, feedback taken

Krishna Sankar December 10, 2008 at 11:49 pm

@Simon : Oh, that means my way of explaining failed.
Let me try again.
It is known that a sine wave and a cosine wave with period T are orthogonal to each other,
i.e. integral_over T sin(wt)*cos(wt) = 0

This means we can send independent information on sine wave and cosine wave and be able to recover both of them at the receiver with out any distortions. This forms the crux of IQ modulation. We send the information I on cosine wave and Q on sine wave.
What is the advantage? Ofcourse – double the data rate.
What is the flipside? We need double the bandwidth. If we send only on sine or cosine, the spectrum is symmetric. We need to send information only on one side of the spectrum. However, with IQ modulation spectrum is no longer symmetric.

Simon December 10, 2008 at 1:04 pm

Krishna,
“Some time back, a friend mentioned that he is yet to understand the need for having both in-phase (I) and quadrature-phase (Q) signals in typical wireless systems”

I wonder if you could please show me some info on this understand, I am still confused why we need I and Q?

Krishna Sankar November 27, 2008 at 5:49 am

If the symbol rate is 1/T and modulation is BPSK, then
(a) bandwidth requirement in baseband is -1/2T to +1/2T.
(b) bandwidth requirement in passband is fc-1/2T to fc OR fc to fc+1/2T

So one would think that the since passband requires only a single sidedd spectrum, I would think that the noise bandwidth is smaller in passband. Do you agree?

Muthukrishnan November 23, 2008 at 12:45 pm

Hi krishna,
The SNR of a transmitted signal in baseband is given by (assuming a binary
or 2PAM)
SNR = (R*Eb) / (B*No/2)
Where R – bit rate and B – bandwidth of baseband signal (i.e., if the
baseband signal is ranging from 0 to 10 KHz then I am taking B= -10 to 10
kHz= 20KHz).
Am i right?
So if I am transmitting the same bit rate in bandpass I need 2B (positive
20 KHz and negative 20 KHz )bandwidth which leads to,
SNR = (R*Eb) / (B*No)
So when I convert a signal from baseband to bandpass with the same bitrate
, the SNR will come down by half. Is this inference correct?
or am I assuming something wrong.
I am learning a lot from your website. More importantly the relation between theory and practice. Its really helpful.
Bye

Krishna Sankar May 17, 2008 at 4:00 pm

Didi May 17, 2008 at 4:46 am

I’m using GNURADIO plattaform to implement SDR for OFDM transceiver.

Yahia April 24, 2008 at 5:16 am

Nice link Krishna. You can see that the book I have recommended is being cited in a number of articles in this link. But as you can concluse from these articles, SDR is about building a configurable transceiver using software.

Krishna Sankar April 23, 2008 at 7:44 am

@Yahia. Thanks for the information.
I just happened to see an SDR Overview article in DSPDesignline.com

Yahia April 23, 2008 at 6:17 am

Here are two definitions of SDR:
1- a radio that is substantially defined in software and whose physical layer behavior can be significantly altered through changes to its software
2- The software radio is a radio which employs one wideband analog to digital converter (ADC) with a resource of programmable digital signal processors (DSPs) to service an entire spectrum allocation in a single integrated module

Now the first one is taken from Jifferey Reed book of Software Radio which I recommend for interested readers. it is titled “Software Radio: A Modern Approach to Radio Engineering ”

To make it clear, SDR is a configurable wireless platform that can change its personality using software “channel frequency, modulation, data rate, transmission power and many others”

Now the ability to sense and “make decisions” to change personality to an appropriate format is called cognitive radio. So in simple words, cognitive radio =SDR+intellegence

tabraiz March 14, 2011 at 4:48 am

What are you doing these days

Krishna Sankar May 24, 2011 at 5:45 am

@tabraiz: idling… try to come back to posting

Krishna Sankar April 23, 2008 at 4:45 am

@Yahia: Yes, it would be interesting to share thoughts.
Quick q: When we say SDR, is it
(a) the RF center frequency which is getting dynamically programmed ?
(b) the baseband processing adapting to various modulation schemes?
OR is it the SDR concept too vast to encapsulate in the above bullets?

Yahia April 22, 2008 at 8:51 am

Unfortunately not Krishna

My work on SDR is part of my research for PhD. But I will be happy to share my work with you. I am using an SDR platform from Lyrtech. At any case I will be happy to discuss any issue about SDR with you. There is a coming standard called 802.22 which will be a cognitive radio based standard. If you are interested, we can exchange ideas and information about these topics.

Krishna Sankar April 22, 2008 at 7:45 am

@Yahia: Thanks.
Do you have a website where information about your work is presented? I would be keen to have a look and get a feel about the work on SDR.

Yahia April 21, 2008 at 10:44 pm

Great job! this website is very impressive. I am working on digital commincations for SDR and I have added this website to my top favorites. Keep it up and thank you for all your efforts.

Somya March 3, 2009 at 11:29 am

Hi Krishna, really a very helpful explanation. Please keep this good work going

Hi Yahia, I had also started working on SDR using GNU Radio platform. If possible, can you please drop me a mail at somya1104@gmail.com so we can discuss our respective works.

Krishna Sankar April 21, 2008 at 7:46 am

@Johny: No worries. I typically do not get disturbed by questions.

jhony April 20, 2008 at 10:27 pm

Hi
yeah thank you very much now its clear… i hope u dont get disturbed by my questions but i really want to clear things up

Krishna Sankar April 20, 2008 at 7:05 am

@johny:
What we want to transmit is
s(t) = aI cos(2*pi*f_ct) – aQsin(2*pi*f_ct)

It is reasonably intuitive to see that above equation is equal to
real part of (aI + jaQ)*(cos(2*pi*f_c*t) + j*sin(2*pi*f_c*t)). Do you agree?

jhony April 20, 2008 at 1:14 am

Hi
thanks… but why do we want to consider only the real part?
thankyou very much for your reply please clear this query. I shall be very thank to you …

Krishna Sankar April 18, 2008 at 8:39 pm

@ Johny: Thanks. Ofcourse, you can shoot in your queries. I will try answer to the best of my knowledge.
The symbol R stands for real, which means the output takes only the real part of s(t) e^{j2*pi*f_c*t).
Helps?

Krishna