I have a question about CDMA, FDMA and TDMA.

I don't understand why we suffer from limited capacity when we have CDMA.

I can understand the limitation in FDMA and TDMA, as time and frequency are limited. However, we can generate as many codes as we want. When the true receiver receives the signal and multiplies by the code, one should receive the original signal even if there is a lot of interference.

So where am I wrong? Why do we suffer from limited capacity when using CDMA?

Thank, I think I made mistake by saying capacity. I specifically meant high date rate.
However, I know multiple access mechanisms.

What I asked is a different question.

Let's say I am using TDMA for 10 users. Each user has to wait 9 users to transmit. If we had 100 users, then each user has to wait 99 users. Very high level of starvation. If you want to decrease starvation, you have to decrease time slot which means decreasing date rate.

Let's say I am using FDMA for 10 users with 100 MHz bandwidth, so each user will have 10 MHz bandwidth so equivalent capacity from Shannon law. If we had 100 users which means 1 MHz bandwidth and much lower date rate equivalently lower capacity. If you want higher date rate, you need to buy more bandwidth.

Let's say I am using CDMA for 10 users, I need 10 orthogonal codes. For 100 users, 100 orthogonal codes.

So, the case is frequency and time are exist in the nature and they are limited.

However, we can generate millions of orthogonal code for millions​ of user as it is human-invented thing.

Why do we live problem multiple access problem​s while we have CDMA?

In other words, is there a mathematical or practical limit of usable number of codes ?

If the number of the users in the network exceeds a threshold, Do the codes become to similar which undermines orthogonality?


2 Answers 2


From an abstract viewpoint, the space $\Omega$ of all signals that are essentially band-limited to a bandwidth of $W$ Hz (say from $f_0$ Hz to $f_0+W$ Hz) and essentially time-limited to duration $T$ (say the epoch lasting from $t=0$ to $t=T$) has approximately $2WT$ orthonormal signals in it, that is, this set $\mathcal B$ of $2WT$ signals is a (orthonormal) basis for this space. Note the words essentially which mean that most of the energy in the signals lies in the indicated bandwidth or epoch. The precise statement of this result (including a description of most etc.) is known as the Landau-Pollak Theorem.

Each signal in the orthonormal basis for $\Omega$ can be used with (say) binary antipodal modulation to transmit one bit, or more bits could be transmitted via $M$-ary amplitude modulation etc (depending on the available transmitter power, the desired error probability, estimated SNR at the receiver). Here, we shall assume for the sake of simplicity and ease of exposition that binary antipodal modulation of the signal must be used. Indeed, all $2WT$ orthonormal basis signals can be used to transmit a total of $2WT$ bits during the epoch $[0,T]$ over this channel of bandwidth $W$, that is, this channel can thus sustain a transmission rate of $2W$ bps. Although the transmissions of the various modulated basis signals all occur within the same epoch and in the same frequency band, matched-filter receivers can separate the individual transmissions because orthonormality means that the sample output of the receiver for the $i$-th basis signal will not contain any interference components from the other basis signals that might be present at the same time.

Now, the orthonormal basis $\mathcal B$ is not unique; many different signal sets that can be used as an orthonormal basis for $\Omega$. Furthermore, it is not necessary that the basis signals all have duration $T$ or occupy the entire bandwidth $W$; it can be advantageous to consider a basis in which the basis signals are of smaller duration or occupy smaller bandwidth, or both. Such choices of orthonormal basis sets gives the schemes known as TDMA and FDMA that allow $K$ users to share the channel of bandwidth $W$

TDMA partitions the epoch $[0,T)$ into $K$ sub-epochs of duration $\frac TK$ with the $i$-th sub-epoch, $1 \leq i \leq K$, being the interval $\left[(i-1)\frac TK, i\frac TK\right)$. Consider the spaces $\Omega^{(i)}$ of signals that are of bandwidth $W$ but are time-limited to the $i$-th sub-epoch, $1 \leq i \leq K$. These are subspaces of $\Omega$ and the Landau-Pollak theorem applied to these subspaces tells us that $\Omega^{(i)}$ has an orthonormal basis $\mathcal B^{(i)}$ of size $2W\frac TK$. (Assume that $2W\frac TK$ is a positive integer). Now, clearly, the space $\Omega$ is the direct sum $$\Omega = \Omega^{(1)} + \Omega^{(2)} + \cdots + \Omega^{(K)}$$ meaning that any signal of bandwidth $W$ and epoch $[0,T)$ can be expressed as the sum of signals in these subspaces. Furthermore, $\bigcup_{=1}^K \mathcal B^{(i)}$ is an orthonormal basis for $\Omega$ since the signals in $\mathcal B^{(i)}$ are orthogonal to each other by the Landau-Pollak theorem and orthogonal to the signals in any other $\mathcal B^{(k)}$ because the signals do not overlap in time. By assigning the $i$-th user the signal set $\mathcal B^{(i)}$, the user communicates $2W\frac TK$ bits in the $\frac TK$-duration time-slot allotted to him, that is, at rate $2W$ bps. Of course, he has to stay silent during the rest of the $T$-second epoch, and so his net transmission rate is $\frac{2W}{K}$ bps, that is, the available transmission rate $2W$ bps is shared equitably among the $K$ users.

A less abstract version is that the $i$-th user is transmitting during $\left[(i-1)\frac TK, i\frac TK\right)$ but can use the full bandwidth $W$ during this short time slot.

FDMA partitions the bandwidth $W$ from $f_0$ to $f_0+W$ Hz into $K$ sub-bands of width $\frac WK$. The $i$-th sub-band occupies the sub-band $\left[f_0 + (i-1)\frac WK, f_0 + i\frac WK\right)$ Hz, and it has $2\frac WK T$ orthonormal signals in it (by the Landau-Pollak theorem applied to the sub-band) where we are assuming, as before, that $2\frac WK T$ is a positive integer. Furthermore, the $2W\frac TK$ signals that span the $i$-th sub-band are orthogonal to the $2W\frac TK$ signals that span the $k$-th sub-band for $k \neq i$. The union of the bases of these sub-bands is the set of $2WT$ basis signals that span $\Omega$. (This abbreviated description can be expanded into one similar to the one for TDMA bu those who prefer to have their i's crossed and their t's dotted wherever possible).

A less abstract version is that the $i$-th user is transmitting in the $i$-th sub-band but can use the full epoch $T$ while being confined to this narrow frequency band. During this transmission, the user sends $2W\frac TK$ bits, and thus has a transmission rate of $\frac{2W}{K}$ bps, that is, the channel rate of $2W$ bps is being shared equitably among the $K$ users.

TDMA and FDMA can be viewed as dividing up the signal space into time-orthogonal and frequency-orthogonal subspaces respectively and assigning a subspace to each user. In the very simplest case with $2WT = 4$ and $K = 4$, the users can be thought of as having been assigned signals as follows:

User #1: $[s_1,0,0,0]$

User #2: $[0,s_1,0,0]$

User #3: $[0,0,s_1,0]$

User #4: $[0,0,0,s_1]$

meaning, for TDMA, that User #1 transmits one bit in time slot #1 using BPSK modulation of $s_1$ (using the full bandwidth of the channel), and remains silent for the remaining three time slots. If you like, you can think of it as modulating that one bit onto $[s_1,0,0,0]$ and transmitting the result. Note that all the energy $\mathscr E_b = \|s_1\|^2$ that User #1 has available for that one bit is concentrated into that one time slot.

CDMA assigns signals as follows.

User #1: $\frac 12[s_1,s_1,s_1,s_1]$

User #2: $\frac 12[s_1,-s_1,s_1,-s_1]$

User #3: $\frac 12[s_1, s_1,-s_1,-s_1]$

User #4: $\frac 12[s_1,-s_1,-s_1,s_1]$

Notice that these signals are orthogonal, just like the previous set!

Now, User #1 is still transmitting just the one bit that he has to transmit over the $T$ second epoch, but he is modulating that bit onto $\frac 12[s_1,s_1,s_1,s_1]$ and transmitting during the entire time. Since he has only energy $\mathscr E_b$ to spend, the amplitude of his transmission is smaller, and he spends energy $$\frac 14 (\|s_1\|^2 +\|s_1\|^2+\|s_1\|^2+\|s_1\|^2) = \|s_1\|^2 = \mathscr E_b$$ just as before. But, in contrast to TDMA or FDMA, each user is transmitting all the time and occupying all the bandwidth (at reduced power), and it is the orthogonality of the coding that is providing the multiple-access capability, not the disjointness in time frequency.

So what does all this have to do with the real concerns of the OP who

  • does not like TDMA because the transmission is bursty,
  • does not like FDMA because each user gets a smaller share of the available bandwidth as the number of users to be accommodated grows,


  • thinks CDMA is the answer because it can accommodate as many users as desired with as high data rates as they desire merely by assigning them orthogonal signals which we can create as orthogonal sequences over the alphabet $\{+1, -1\}$ or $\{(-1)^0, (-1)^1\}$ for those who think of binary sequences in terms of zeros and ones?

Well, as Marcus Muller's answer points out more succinctly than the prolix blabbering above, linear algebra tells us that we can find no more than $N$ orthogonal vectors in a vector space of dimension $N$. 10 users can share a 100 MHz channel (with $200\times 10^6$ orthogonal signals per second in it) equitably via TDMA or FDMA or CDMA, but if 100 users have to be accommodated, then each can only get $2\times 10^6$ orthogonal signals for use; it is not possible to get them the silk purse of $20\times 10^6$ orthogonal signals each from the sow's ear of $100$ MHz bandwidth regardless of whether we use TDMA or FDMA or CDMA. From a more practical perspective, consider that if a fixed data bit duration is divided into $100$ "chips" instead of just $10$ chips so that we can accommodate $100$ users via CDMA instead of just $10$ users (all having the same data rate), then the bandwidth is perforce increased by a factor of $100$ instead of a factor of $10$. Thus, CDMA cannot "manufacture" additional capacity by using a man-made construct of bit sequences that does not have the limitations of the natural notions of frequency and time.

  • 1
    $\begingroup$ This is greatly summarises what i tried to mean.So i am waiting CDMA part $\endgroup$
    – Mio
    Commented Apr 17, 2017 at 11:42
  • $\begingroup$ Nice to see a question provoke such an extensive answer from your side, Dilip :) This is impressive work you put into this; I'd actually like to see your answer being accepted, especially since it also addresses the the non-uniform medium "splitting" correctly (would've become really awkward to somehow manipulate that into my answer). Pretty sure this would become a "go-to" answer in the future :) $\endgroup$ Commented Apr 19, 2017 at 21:26

You're misunderstanding what CDMA, TDMA and FDMA do:

CDMA doesn't increase the channel capacity in any way. It's a MA, MA = Multiple Access mechanism.

In other words, it's just a way of dividing the spectrum among multiple users. No matter what you do, you can't get more data through a channel than the physics and math allows – and Shannon's channel capacity gives a pretty clear upper boundary for that.

All the mechanisms you quote hence only divide the available channel across some measure.

For TDMA and FDMA, as you notice, this is very intuitively understandable.

But to be honest, CDMA is not that different. Instead of sending a single symbol in a single symbol time slot, you have to multiply it onto a longer code sequence.

The maximum number of simultaneous users of CDMA is simply the number $N$ of orthogonal CDMA sequences.

$N$ orthogonal vectors (i.e. sequences) form a base to an $N$-dimensional vector space (which is the signal space). Vectors from an $N$-dimensional space (which is what you're going to transmit to transport one symbol) have inherently $N$ elements.

So, to make (up to) $N$ people talk at once, you need to "blow up" everyone's transmission's symbols (which are $1$-dimensional, each) to $N$-dimensional vectors, increasing the time it takes to send one symbol by a factor of $N$.

That's the same efficiency as using TDMA to split time into $N$ slots (everyone sends $1$ symbol every $N$ slots), or dividing the bandwidth into $N$ channels and using FDMA (everyone can only use $\frac1N$ of the available bandwidth, reducing the individual symbol rate by $N$).

  • 1
    $\begingroup$ Note that, from an information-theoretical standpoint, the capacity (b/s/Hz) of CDMA is smaller than that of TDMA/FDMA (which are equivalent). When taking into account practical constraints (guard bands, guard intervals, preamble overhead, etc), the tree approaches yield different capacities. $\endgroup$
    – vaz
    Commented Apr 19, 2017 at 21:04

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