Frequency-Shift Keying (FSK)

Frequency-Shift Keying (FSK) is a digital modulation technique in which the frequency of a carrier signal is varied to represent distinct symbol states. Each symbol is associated with a specific (sub) carrier frequency.

FSK Waveform

The FSK signal can be expressed as

\[ \boxed{ s_m(t) = \sqrt{\frac{2\mathcal{E}}{T}} \cos\bigl(2\pi f_c t + 2\pi m \Delta f t \bigr),} \quad 1 \le m \le M,\quad 0 \le t \le T \]

where:

• \(\mathcal{E}\) is the signal energy,

• \(T\) is the symbol duration,

• \(f_c\) is the carrier frequency,

• \(\Delta f\) is the frequency separation between successive signals,

• \(m\) indexes the different frequencies.

The lowpass equivalent of \(s_m(t)\), denoted \(s_{m,lp}(t)\), is given by

\[ \boxed{ s_{m,lp}(t) = \sqrt{\frac{2\mathcal{E}}{T}} e^{j 2\pi m \Delta f t} } \]

Note that, \(s_m(t)\) is a real-valued, bandpass waveform (a cosine at frequency \(f_c + m \Delta f\)).

By contrast, the lowpass-equivalent signal \(s_{m,lp}(t)\) is complex-valued, capturing the in-phase and quadrature components of the bandpass signal at baseband.

Orthogonality of FSK Signals

Correlation of Two Signals

The correlation of the lowpass equivalents \(s_{m,lp}(t)\) and \(s_{n,lp}(t)\) is calculated as

\[ \langle s_{m,lp}(t), s_{n,lp}(t)\rangle = \int_0^T s_{m,lp}(t) s_{n,lp}^*(t) \mathrm{d}t = \frac{2\mathcal{E}}{T}\int_{0}^{T} e^{ j 2\pi (m-n) \Delta f t} \mathrm{d}t \]

Evaluating this integral yields:

\[ \langle s_{m,lp}(t), s_{n,lp}(t)\rangle = \frac{2\mathcal{E}}{T} \frac{\sin \bigl(\pi T (m-n) \Delta f\bigr)} {\pi (m-n) \Delta f} \mathcal{E}^{ j \pi T (m-n) \Delta f} \]

Taking its real part, we have

\[ \boxed{ \mathrm{Re} \bigl[\langle s_{m,lp}(t),s_{n,lp}(t)\rangle\bigr] = \frac{2\mathcal{E}}{T} \mathrm{sinc} \bigl(2 T (m-n) \Delta f\bigr) } \]

where \(\mathrm{sinc}(x) \equiv \dfrac{\sin(\pi x)}{\pi x}\).

Proof: See Appendix.

Condition for Orthogonality

For orthogonality, we need this inner product to be zero for \(m \neq n\). Hence,

\[ \mathrm{Re} \bigl[\langle s_{m,lp}(t), s_{n,lp}(t)\rangle\bigr] = \frac{2\mathcal{E}}{T} \mathrm{sinc} \bigl(2 T (m-n) \Delta f\bigr) = 0 \]

Since the nonzero constant \(\tfrac{2\mathcal{E}}{T}\) just scales the expression, the condition boils down to

\[ \mathrm{sinc} \bigl(2 T (m-n) \Delta f\bigr) = 0, \quad \text{for all } m \neq n \]

Zeros of the Sinc Function

Recall that

\[ \mathrm{sinc}(x) = \frac{\sin(\pi x)}{\pi x} \]

has zeros at all nonzero integer values of \(x\), i.e.,

\[ x = \pm 1, \pm 2, \pm 3,\dots \]

Hence

\[ 2 T (m-n) \Delta f = k \quad \text{for } k \in \{\pm1, \pm2, \pm3, \ldots\}, k \neq 0 \]

Minimum Frequency Separation

Adjacent Frequencies:
For \(M\)-FSK, the frequencies are typically spaced in increments of \(\Delta f\). Adjacent signals correspond to \((m-n)=1\). So the simplest (lowest nonzero) orthogonality condition is

\[ 2 T (m-n) \Delta f = 2 T \Delta f = k=1 \quad\Longrightarrow\quad \Delta f = \frac{1}{2T} \]

In other words, for adjacent frequency indices \(m\) and \(m+1\), the separation must be

\[ \Delta f = \frac{1}{2T} \]

(if we choose \(k=1\) for the first zero of \(\mathrm{sinc}\)).

General \(k\)-multiple:
More generally, we could choose

\[ 2 T \Delta f = k \quad\Longrightarrow\quad \Delta f = \frac{k}{2T}, \quad k=1,2,3,\dots \]

But the minimum spacing that still guarantees orthogonality for adjacent channels is with \(k=1\). This is why we say

\[ \boxed{ \Delta f_{\min} = \frac{1}{2T} } \]

Orthogonality in Baseband and Bandpass

Because the bandpass signals \(s_{m}(t)\) differ by frequency increments of \(\Delta f\), their lowpass equivalents differ by the same increments in baseband (just “shifted” from the carrier). The condition

\[ 2 T (m-n) \Delta f \in \mathbb{Z} \]

ensures pairwise orthogonality of \(\{s_{m,lp}(t)\}\). It equally implies the real bandpass signals \(\{s_{m}(t)\}\) are orthogonal (their integrals over one symbol period vanish). Thus, no symbol‐to‐symbol interference occurs when \(\Delta f\) meets or exceeds that minimum value.

Energy of FSK Waveform

When \(m = n\), we are effectively taking the autocorrelation of the same signal \(s_{m,lp}(t)\) with itself. In that case, the exponential term inside the integral becomes

\[ e^{j 2\pi (m-n) \Delta f t} = \mathcal{E}^{j 2\pi (0) \Delta f t} = 1 \]

so the integral simplifies to

\[ \int_{0}^{T} 1 \mathrm{d}t = T \]

Given the factor \(\frac{2\mathcal{E}}{T}\) out front, the correlation becomes

\[ \langle s_{m,lp}(t), s_{m,lp}(t)\rangle = \frac{2\mathcal{E}}{T} \times T = 2\mathcal{E} \]

Thus, when \(m=n\),

• The correlation is nonzero and equals the signal energy \(2\mathcal{E}\).

• This makes sense physically: the energy of the signal over one symbol interval is \(2\mathcal{E}\) (based on the factor \(\sqrt{\tfrac{2\mathcal{E}}{T}}\) in the signal definition).

Hence, orthogonality pertains only to the case \(m\neq n\). When \(m=n\), the “correlation with itself” is simply the (nonzero) autocorrelation/energy of that signal.

Direct Derivation of the Energy of the Bandpass Waveform

Given Definitions

• Bandpass FSK signal:

  \[ s_m(t) = \sqrt{\frac{2\mathcal{E}}{T}} \cos \bigl(2\pi f_c t + 2\pi m \Delta f t\bigr), \quad 0 \le t \le T \]

• Complex baseband equivalent:

  \[ s_{m,lp}(t) = \sqrt{\frac{2\mathcal{E}}{T}} e^{ j 2\pi m \Delta f t} \]

Energy of \(s_{m,lp}(t)\)
The energy of the complex lowpass signal \(s_{m,lp}(t)\) is

\[ \int_{0}^{T} \bigl|s_{m,lp}(t)\bigr|^2 dt = \int_{0}^{T} \left(\sqrt{\frac{2\mathcal{E}}{T}}\right)^2 dt = \int_{0}^{T} \frac{2\mathcal{E}}{T} \mathrm{d}t = 2\mathcal{E} \]

Thus, its complex representation has total energy \(2\mathcal{E}\).

Energy of \(s_m(t)\)
The real bandpass signal is

\[ s_m(t) = \sqrt{\frac{2\mathcal{E}}{T}} \cos \bigl(\omega t + \phi\bigr) \]

where \(\omega = 2\pi (f_c + m \Delta f)\).

To find its energy over \(t \in [0, T]\):

\[ \int_{0}^{T} s_m^2(t) \mathrm{d}t = \int_{0}^{T} \left(\sqrt{\frac{2\mathcal{E}}{T}}\right)^2 \cos^2(\omega t + \phi) \mathrm{d}t \]

Inside the integral, \(\cos^2(\omega t + \phi)\) has an average value of \(1/2\) over an integer number of cycles. For FSK over one symbol interval \(T\), we typically choose \(\omega\) (or \(\Delta f\)) so that the signal completes an integer number of half‐cycles or full cycles, making

\[ \int_0^T \cos^2(\omega t + \phi) dt = \frac{T}{2} \]

Therefore,

\[ \int_{0}^{T} s_m^2(t) dt = \int_{0}^{T} \frac{2\mathcal{E}}{T} \cdot \cos^2(\omega t + \phi) \mathrm{d}t = \frac{2\mathcal{E}}{T} \times \frac{T}{2} = \mathcal{E} \]

So the passband signal \(s_m(t)\) has total energy \(\mathcal{E}\).

Interpretation

• The baseband equivalent signal \(s_{m,lp}(t)\) is a \(\textit{complex}\)-valued representation: it keeps track of the in-phase (I) and quadrature (Q) components. In this representation, the “energy” is found by integrating \(\lvert s_{m,lp}(t)\rvert^2\), and turns out to be \(2\mathcal{E}\).

• The actual physical (bandpass) signal \(s_m(t)\) is \(\textit{real}\). Because the complex representation effectively accounts for two orthogonal dimensions (I and Q), it naturally has twice the energy when viewed as a complex 2D vector. In the real bandpass world, we only send one dimension (the cosine), so the integral of \(s_m^2(t)\) over the symbol interval is \(\mathcal{E}\).

Hence, seeing a factor of 2 difference between the energies of the complex baseband signal and the real passband signal is completely normal in digital communications.

Orthonormal Basic for FSK

We construct an orthonormal basis for orthogonal M‐FSK waveforms and then represent each FSK signal as a point in an \(M\)‐dimensional Euclidean space.

We assume that the frequency separation \(\Delta f\) is chosen so that the signals \( \{s_m(t)\} \) are pairwise orthogonal over the symbol interval \([0, T]\). In particular, the minimum spacing ensuring orthogonality of adjacent frequencies is

\[ \Delta f = \frac{1}{2T} \]

Definition of the FSK Signals

We start with the real, bandpass M‐FSK signals:

\[ s_m(t) = \sqrt{\frac{2\mathcal{E}}{T}} \cos \Bigl(2\pi f_c t + 2\pi m \Delta f t\Bigr), \quad m=1,2,\dots, M, \quad 0 \le t \le T \]

where:

• \(\mathcal{E}\) is the energy parameter used in normalizing the signals,

• \(T\) is the symbol duration,

• \(f_c\) is the carrier frequency,

• \(\Delta f\) is the frequency separation between adjacent FSK tones (typically \(\Delta f = \tfrac{1}{2T}\) for minimum orthogonal spacing).

Constructing the Orthonormal Basis \(\{\phi_k(t)\}\)

To form an orthonormal set of basis functions that spans all \(M\) FSK signals, define for \(k = 1,2,\dots, M\):

\[ \phi_k(t) = \sqrt{\frac{2}{T}} \cos \Bigl(2\pi f_c t + 2\pi (k-1) \Delta f t\Bigr), \quad 0 \le t \le T \]

Orthogonality

Consider the inner product over one symbol interval,

\[ \langle \phi_k, \phi_\ell \rangle = \int_{0}^{T} \phi_k(t) \phi_\ell(t) \mathrm{d}t \]

By design—assuming \(\Delta f\) is chosen such that different tones are orthogonal—these basis functions satisfy

\[ \langle \phi_k, \phi_\ell \rangle = \delta_{k\ell} \]

where \(\delta_{k\ell}\) is the Kronecker delta (1 if \(k=\ell\), 0 otherwise).

Normalization

Each \(\phi_k(t)\) has unit energy:

\[ \int_0^T \phi_k^2(t) \mathrm{d}t = 1 \]

(This follows from the factor \(\sqrt{2/T}\) in front of a \(\cos^2\) function, which integrates to \(T/2\) over the interval.)

Hence, \(\{\phi_k(t)\}_{k=1}^M\) is an orthonormal set in the real signal space.

Expressing \(s_m(t)\) in Terms of \(\{\phi_k(t)\}\)

Observe that

\[ s_m(t) = \sqrt{\frac{2\mathcal{E}}{T}} \cos \Bigl(2\pi f_c t + 2\pi m \Delta f t\Bigr) \]

Compare this with

\[ \phi_m(t) = \sqrt{\frac{2}{T}} \cos \Bigl(2\pi f_c t + 2\pi (m-1) \Delta f t\Bigr) \]

If we choose the index consistently (say we shift \(m\to m-1\) in the definition of \(\phi\) or vice versa), each \(s_m(t)\) differs from its corresponding \(\phi_m(t)\) only by an amplitude factor \(\sqrt{\mathcal{E}}\). Indeed:

\[ \phi_m(t) = \sqrt{\frac{2}{T}} \cos \Bigl(2\pi f_c t + 2\pi (m-1) \Delta f t\Bigr) \]

But if we re‐label or if we define

\[ s_{m}(t) = \sqrt{\mathcal{E}} \phi_{m+1}(t) \]

(depending on indexing), we get exactly the same waveform shape with amplitude \(\sqrt{2\mathcal{E}/T}\). The key takeaway is:

\[ \boxed{ s_m(t) = \sqrt{\mathcal{E}} \phi_m(t)} \quad \text{(up to a small frequency‐index shift)} \]

Hence, each FSK signal \(s_m(t)\) is essentially \(\sqrt{\mathcal{E}}\) times one of the orthonormal basis functions \(\phi_m(t)\).

FSK Signal Space

Mapping to an \(M\)‐Dimensional Euclidean Space

Once we have an orthonormal set \(\{\phi_k(t)\}_{k=1}^M\), any signal \(x(t)\) living in the span can be written uniquely as:

\[ x(t) = \sum_{k=1}^M \alpha_k \phi_k(t) \]

• The coordinate of \(x(t)\) in this \(M\)‐dimensional space is
  \(\bigl(\alpha_1, \alpha_2, \dots, \alpha_M\bigr)\).

• If \(x(t) = s_m(t)\), then all but the \(m\)th coordinate will be zero, and that \(m\)th coordinate will be \(\sqrt{\mathcal{E}}\). Symbolically,

  \[ s_1(t) \longmapsto (\sqrt{\mathcal{E}}, 0, 0,\dots, 0) \]
  \[ s_2(t) \longmapsto (0, \sqrt{\mathcal{E}}, 0,\dots, 0) \]
  \[ \quad \dots \quad \]
  \[ s_M(t) \longmapsto (0, 0, \dots, 0, \sqrt{\mathcal{E}}) \]

Therefore, orthogonal M‐FSK can be viewed geometrically as placing each symbol at a vertex of an \(M\)‐dimensional simplex whose edges have length \(\sqrt{2 \mathcal{E}}\). Each symbol waveform is “one basis function turned on” (multiplied by \(\sqrt{\mathcal{E}}\)).

Euclidean and Minimum Distance

We compute the pairwise Euclidean distance between FSK signals (assuming they are orthogonal) and determine the minimum distance.

Definition of the FSK Signals

Consider an \(M\)-ary FSK system with signals:

\[ s_m(t) = \sqrt{\frac{2\mathcal{E}}{T}} \cos \Bigl(2\pi \bigl(f_c + m \Delta f\bigr) t\Bigr), \quad m = 1, 2, \ldots, M, \quad 0 \le t \le T \]

Each signal has energy

\[ \int_{0}^{T} s_m^2(t) \mathrm{d}t = \mathcal{E} \]

(As discussed previously, the factor \(\sqrt{\tfrac{2\mathcal{E}}{T}}\) in front of a cosine means each real passband signal integrates to \(\mathcal{E}\) over the interval \([0,T]\).)

We assume \(\Delta f\) is chosen so that the signals \(\{s_m(t)\}\) are orthogonal—i.e.,

\[ \int_{0}^{T} s_m(t) s_n(t) \mathrm{d}t = 0, \quad m \neq n \]

Euclidean Distance Between Two FSK Signals

In signal space, the squared Euclidean distance between two distinct signals \(s_m(t)\) and \(s_n(t)\) is given by

\[ d^2(m,n) = \int_{0}^{T} \Bigl[s_m(t) - s_n(t)\Bigr]^{2} \mathrm{d}t \]

Expanding this:

\[ d^2(m,n) = \int_{0}^{T} \Bigl[s_m^2(t) - 2 s_m(t) s_n(t) + s_n^2(t)\Bigr] \mathrm{d}t \]

Using orthogonality, we have:

\[\displaystyle \int_{0}^{T} s_m^2(t) \mathrm{d}t = \mathcal{E}\]

\[ \displaystyle \int_{0}^{T} s_n^2(t) \mathrm{d}t = \mathcal{E}\]

\[ \displaystyle \int_{0}^{T} s_m(t) s_n(t) \mathrm{d}t = 0\quad \text{for }m \neq n\]

Hence,

\[ d^2(m,n) = \mathcal{E} + \mathcal{E} - 2 \times 0 = 2 \mathcal{E} \]

Therefore,

\[ d(m,n) = \sqrt{2 \mathcal{E}} \]

Minimum Distance

Because all pairs of distinct signals are orthogonal and have the same energy \(\mathcal{E}\), their pairwise distances are all the same, namely \(\sqrt{2 \mathcal{E}}\). Therefore, the minimum distance (among all possible pairs) is

\[ \boxed{ d_{\min} = \sqrt{2 \mathcal{E}} } \]

In an \(M\)-FSK constellation (with ideal orthogonality), every pair of signals is separated by the same distance \(\sqrt{2 \mathcal{E}}\).

Geometric Interpretation

When mapped into an \(M\)-dimensional orthonormal basis (each basis vector corresponding to one frequency tone), the signals \(s_m(t)\) simply become vectors whose coordinates are all zero except for the \(m\)th entry, which is \(\sqrt{\mathcal{E}}\). For example:

\[ s_1(t) \longmapsto (\sqrt{\mathcal{E}}, 0, 0, \dots) \]

\[ s_2(t) \longmapsto (0, \sqrt{\mathcal{E}}, 0, \dots) \]

\[ \vdots \]

\[ s_M(t) \longmapsto (0, 0, \dots, \sqrt{\mathcal{E}}) \]

Each vector thus has length \(\sqrt{\mathcal{E}}\). The distance between any two distinct vertices in this orthonormal “simplex” is

\[ \sqrt{(\sqrt{\mathcal{E}} - 0)^2 + (0 - \sqrt{\mathcal{E}})^2 + \dots} = \sqrt{\mathcal{E} + \mathcal{E}} = \sqrt{2 \mathcal{E}} \]

Minimum Distance in terms of \(\mathcal{E}_b\)

The minimum Euclidean distance of orthogonal M‐FSK signals can be expressed in terms of the energy per bit \(\mathcal{E}_b\) as follows.

Relationship Between \(\mathcal{E}_s\) (Energy per Symbol) and \(\mathcal{E}_b\) (Energy per Bit)

For an \(M\)‐ary scheme, each symbol carries \(\log_2(M)\) bits. If \(\mathcal{E}_s\) denotes the energy to send one symbol, and \(\mathcal{E}_b\) denotes the energy to send one bit, the relationship is:

\[ \mathcal{E}_s = (\log_2 M) \mathcal{E}_b \]

This simply says that the energy for a symbol of \(\log_2(M)\) bits is \(\log_2(M)\) times the energy per bit.

Minimum Distance of Orthogonal M‐FSK in Terms of \(\mathcal{E}_s\)

For orthogonal M‐FSK, each signal (of duration \(T\)) has energy \(\mathcal{E}_s\). The pairwise Euclidean distance \(d(m,n)\) between any two distinct signals \(s_m(t)\) and \(s_n(t)\) is

\[ d(m,n) = \sqrt{ \int_0^T \bigl[s_m(t) - s_n(t)\bigr]^2 \mathrm{d}t } = \sqrt{ 2 \mathcal{E}_s } \]

Because all signals are mutually orthogonal, all pairs have the same distance \(\sqrt{2\mathcal{E}_s}\). Hence,

\[ d_{\min} = \sqrt{2 \mathcal{E}_s} \]

Expressing \(d_{\min}\) in Terms of \(\mathcal{E}_b\)

Using \(\mathcal{E}_s = (\log_2 M) \mathcal{E}_b\), substitute into \(d_{\min} = \sqrt{2 \mathcal{E}_s}\):

\[ d_{\min} = \sqrt{2 (\log_2 M) \mathcal{E}_b} = \sqrt{2 \log_2(M)} \sqrt{\mathcal{E}_b} \]

Therefore,

\[ \boxed{ d_{\min} = \sqrt{2 \log_2(M) \mathcal{E}_b} } \]

Linear vs. Nonlinear Modulation

• In linear modulation schemes (e.g., ASK, PSK, QAM), the sum of two signals is still a valid signal within the same modulation scheme.

• In nonlinear modulation schemes like FSK, this is not true because the frequencies used in FSK are distinct, and the sum of two FSK signals does not belong to the same scheme.

Appendix: Derivation of The Correlation of Two FSK Signals

We have two lowpass signals:

\[ s_{m,lp}(t) = \sqrt{\frac{2\mathcal{E}}{T}} e^{ j 2\pi m \Delta f t}, \quad s_{n,lp}(t) = \sqrt{\frac{2\mathcal{E}}{T}} e^{ j 2\pi n \Delta f t} \]

The correlation (inner product) over \(0\le t \le T\) is defined as

\[ \langle s_{m,lp}(t), s_{n,lp}(t)\rangle = \int_{0}^{T} s_{m,lp}(t) \bigl(s_{n,lp}(t)\bigr)^{*} \mathrm{d}t \]

Since \(\bigl(s_{n,lp}(t)\bigr)^{*} = \sqrt{\tfrac{2\mathcal{E}}{T}} e^{- j 2\pi n \Delta f t},\)
we get

\[ \langle s_{m,lp}(t), s_{n,lp}(t)\rangle = \int_{0}^{T} \sqrt{\frac{2\mathcal{E}}{T}} e^{j 2\pi m \Delta f t} \sqrt{\frac{2\mathcal{E}}{T}} e^{- j 2\pi n \Delta f t} \mathrm{d}t = \frac{2\mathcal{E}}{T}\int_0^T e^{j 2\pi (m-n) \Delta f t} \mathrm{d}t \]

Thus, we have

\[ \langle s_{m,lp}(t), s_{n,lp}(t)\rangle = \frac{2\mathcal{E}}{T}\int_0^T e^{j 2\pi (m-n) \Delta f t} \mathrm{d}t \]

Next, we evaluate

\[ I = \int_0^T e^{j 2\pi (m-n) \Delta f t} \mathrm{d}t = \left. \frac{ \mathcal{E}^{ j 2\pi (m-n) \Delta f t} }{ j 2\pi (m-n) \Delta f } \right\rvert_{ 0}^{ T} \]

That is,

\[ I = \frac{ \mathcal{E}^{ j 2\pi (m-n) \Delta f T} - 1 } { j 2\pi (m-n) \Delta f } \]

A common way to rewrite \(e^{j\alpha} - 1\) is

\[ e^{j\alpha} - 1 = e^{ j \tfrac{\alpha}{2}} \Bigl(e^{ j \tfrac{\alpha}{2}} - e^{- j \tfrac{\alpha}{2}}\Bigr) = 2 j \mathcal{E}^{ j \tfrac{\alpha}{2}} \sin \Bigl(\tfrac{\alpha}{2}\Bigr) \]

Let \(\alpha = 2\pi (m-n)\Delta f T.\) Then

\[ e^{j\alpha} - 1 = 2 j \mathcal{E}^{ j \alpha/2} \sin \Bigl(\tfrac{\alpha}{2}\Bigr) = 2 j \mathcal{E}^{ j \pi T (m-n) \Delta f} \sin\Bigl(\pi T (m-n) \Delta f\Bigr) \]

Hence,

\[\begin{split} \begin{align*} I &= \frac{ 2 j \mathcal{E}^{ j \pi T (m-n) \Delta f} \sin\bigl( \pi T (m-n) \Delta f\bigr) } { j 2\pi (m-n) \Delta f } \\ &= \frac{ \mathcal{E}^{ j \pi T (m-n) \Delta f} \sin\bigl( \pi T (m-n) \Delta f\bigr) } { \pi (m-n) \Delta f } \end{align*} \end{split}\]

Consequently,

\[ \langle s_{m,lp}(t), s_{n,lp}(t)\rangle = \frac{2\mathcal{E}}{T} \frac{\sin\bigl( \pi T (m-n) \Delta f\bigr) \mathcal{E}^{ j \pi T (m-n) \Delta f}} {\pi (m-n) \Delta f} \]

Thus, we have

\[ \langle s_{m,lp}(t), s_{n,lp}(t)\rangle = \frac{2\mathcal{E}}{T} \frac{\sin\bigl(\pi T (m-n)\Delta f\bigr)}{\pi (m-n)\Delta f} e^{ j \pi T (m-n) \Delta f} \]

Real Part of the Correlation

We now want $$ \mathrm{Re} \Bigl[\langle s_{m,lp}(t), s_{n,lp}(t)\rangle\Bigr]

\mathrm{Re} \biggl[ \frac{2\mathcal{E}}{T} \frac{\sin\bigl( \pi T (m-n) \Delta f\bigr)} {\pi (m-n) \Delta f} e^{ j \pi T (m-n) \Delta f} \biggr]. $$

Let \(\theta = \pi T (m-n) \Delta f\). Then

\[ \sin(\theta) \mathcal{E}^{ j \theta} = \sin(\theta) \bigl[\cos(\theta) + j \sin(\theta)\bigr] = \sin(\theta) \cos(\theta) + j \sin^2(\theta) \]

Hence, the real part is

\[ \sin(\theta) \cos(\theta) = \tfrac12 \sin\bigl(2\theta\bigr) \quad\text{(since }\sin(\theta)\cos(\theta) = \tfrac12[\sin(\theta+\theta) + \sin(\theta-\theta)] = \tfrac12 \sin(2\theta)\bigl) \]

Therefore,

\[ \mathrm{Re} \bigl[\langle s_{m,lp}(t), s_{n,lp}(t)\rangle\bigr] = \frac{2\mathcal{E}}{T} \frac{1}{\pi (m-n) \Delta f} \sin(\theta) \cos(\theta) = \frac{2\mathcal{E}}{T} \frac{1}{\pi (m-n) \Delta f} \frac12 \sin\bigl(2\theta\bigr) \]

Simplify:

\[ = \frac{\mathcal{E}}{T} \frac{\sin\bigl(2 \theta\bigr)}{\pi (m-n) \Delta f} = \frac{\mathcal{E}}{T} \frac{\sin\Bigl(2 \pi T (m-n) \Delta f\Bigr)} {\pi (m-n) \Delta f} \]

Expressing in Terms of \(\mathrm{sinc}(\cdot)\)

Recall the standard definition:

\[ \mathrm{sinc}(x) = \frac{\sin(\pi x)}{\pi x} \]

In our case,

\[ \sin\bigl(2\pi x\bigr) = 2\pi x \mathrm{sinc}(2x) \times\text{(up to factors of 2)} \]

More precisely,

\[ \frac{\sin\bigl(2\pi x\bigr)}{\pi x} = 2 \mathrm{sinc}\bigl(2x\bigr) \]

Hence if we let \(x = T (m-n) \Delta f\), then

\[ \frac{\sin\bigl(2\pi T (m-n) \Delta f\bigr)}{\pi (m-n) \Delta f} = 2 \mathrm{sinc}\bigl(2 T (m-n) \Delta f\bigr) \]

Therefore,

\[ \mathrm{Re} \bigl[\langle s_{m,lp}(t), s_{n,lp}(t)\rangle\bigr] = \frac{\mathcal{E}}{T} \times 2 \mathrm{sinc} \Bigl(2 T (m-n) \Delta f\Bigr) = \frac{2\mathcal{E}}{T} \mathrm{sinc} \Bigl(2 T (m-n) \Delta f\Bigr) \]

Thus, we have

\[ \boxed{ \mathrm{Re} \bigl[\langle s_{m,lp}(t), s_{n,lp}(t)\rangle \bigr] = \frac{2\mathcal{E}}{T} \mathrm{sinc} \Bigl(2 T (m-n) \Delta f\Bigr) } \]