The amplifier you linked to is not a signal generator. It cannot itself generate a signal, variable frequency or otherwise. It is a regular power amplifier, able to deliver up to 260W of power into one or more loudspeakers, but you will still need a signal generator to produce the signal for this amplifier to amplify.

The 80Hz to 15kHz specification tells you what range of frequencies it can amplify without excessive signal attenuation. That's pretty poor for an audio amplifier, so I suspect this one is intended for public address. There's no mention that I can find of how many speakers it is designed to drive, or what impedance those speakers should have. The figure of 260W could be a total for multiple speakers, or a power rating per speaker. I suspect it's the former.

You don't say what impedance your speaker has. I'll assume 4Ω, though common values are 2Ω or 8Ω. You also don't tell us what shape of signal you intend to use, so I'll assume it's a sinusoid.

To deliver 1kW of power to a 4Ω load, if we were talking about DC, would require a certain voltage, which can be calculated using one of the variants of power law \$P=\frac{V^2}{R}\$, which gives us \$V_{DC}\$:

$$ \begin{aligned} V_{DC} &= \sqrt{PR} \\ \\ &= \sqrt{1kW \times 4\Omega} \\ \\ &= 63V \end{aligned} $$

This would correspond to a certain current, which is calculated using Ohm's law \$V=I\times R\$:

$$ \begin{aligned} I_{DC} &= \frac{V_{DC}}{R} \\ \\ &= \frac{63V}{4\Omega} \\ \\ &= 16A \end{aligned} $$

For AC, which is what you will be using to produce audio, we usually work in terms of RMS (root-mean-square), which is the equivalent DC that would deliver the same power to a resistive load by an AC source. For instance, 220V AC RMS from the mains will deliver the same power (to a resistive load) as a 220V DC battery would.

For sinusoids, their peak value is \$\sqrt{2}\$ times greater than a steady DC equivalent, so to calculate the peak values of a sinusoidal voltage source that would deliver the same power:

$$ \begin{aligned} V_{AC(PEAK)} &= V_{RMS} \times \sqrt{2} \\ \\ &= 63V \times 1.4 \\ \\ &= 88V \end{aligned} $$

The sinusoidal AC voltage you apply across your 4Ω speaker, to deliver 1kW of power, will rise to a maximum of +88V, then fall through zero continuing to −88V, and so on.

The corresponding current will be:

$$ \begin{aligned} I_{AC(PEAK)} &= I_{RMS} \times \sqrt{2} \\ \\ &= 16A \times 1.4 \\ \\ &= 22A \end{aligned} $$

There are a hundred ways to produce this kind of source, all of which are difficult. You are certainly better off buying off the shelf, and since home audio almost never calls for such power, you'll probably be looking for a public address system.

1kW of audio power can deafen you, literally. Be cautious, and stand many many metres away from any audio source of that power before switching it on.

The amplifier you linked to is not a signal generator. It cannot itself generate a signal, variable frequency or otherwise. It is a regular power amplifier, able to deliver up to 260W of power into one or more loudspeakers, but you will still need a signal generator to produce the signal for this amplifier to amplify.

The 80Hz to 15kHz specification tells you what range of frequencies it can amplify without excessive signal attenuation. That's pretty poor for an audio amplifier, so I suspect this one is intended for public address. There's no mention that I can find of how many speakers it is designed to drive, or what impedance those speakers should have. The figure of 260W could be a total for multiple speakers, or a power rating per speaker. I suspect it's the former.

You don't say what impedance your speaker has. I'll assume 4Ω, though common values are 2Ω or 8Ω. You also don't tell us what shape of signal you intend to use, so I'll assume it's a sinusoid.

To deliver 1kW of power to a 4Ω load, if we were talking about DC, would require a certain voltage, which can be calculated using one of the variants of power law \$P=\frac{V^2}{R}\$, which gives us \$V_{DC}\$:

$$ \begin{aligned} V_{DC} &= \sqrt{PR} \\ \\ &= \sqrt{1kW \times 4\Omega} \\ \\ &= 63V \end{aligned} $$

This would correspond to a certain current, which is calculated using Ohm's law \$V=I\times R\$:

$$ \begin{aligned} I_{DC} &= \frac{V_{DC}}{R} \\ \\ &= \frac{63V}{4\Omega} \\ \\ &= 16A \end{aligned} $$

For AC, which is what you will be using to produce audio, we usually work in terms of RMS (root-mean-square), which is the equivalent DC that would deliver the same power to a resistive load by an AC source. For instance, 220V AC RMS from the mains will deliver the same power (to a resistive load) as a 220V DC battery would.

For sinusoids, their peak value is \$\sqrt{2}\$ times greater than a steady DC equivalent, so to calculate the peak values of a sinusoidal voltage source that would deliver the same power:

$$ \begin{aligned} V_{AC(PEAK)} &= V_{RMS} \times \sqrt{2} \\ \\ &= 63V \times 1.4 \\ \\ &= 88V \end{aligned} $$

The sinusoidal AC voltage you apply across your 4Ω speaker, to deliver 1kW of power, will rise to a maximum of +88V, then fall through zero continuing to −88V, and so on.

The corresponding current will be:

$$ \begin{aligned} I_{AC(PEAK)} &= I_{RMS} \times \sqrt{2} \\ \\ &= 16A \times 1.4 \\ \\ &= 22A \end{aligned} $$

There are a hundred ways to produce this kind of source, all of which are difficult. You are certainly better off buying off the shelf, and since home audio almost never calls for such power, you'll probably be looking for a public address system.

There are plenty of cheap audio waveform generators on EBay and Amazon and such, most of which will produce the signal you require, to be amplified. You could even use a PC or smartphone to get a good enough sinusoid (from the headphone or line-out socket), though you'll need to measure the signal amplitude yourself, using an RMS AC voltmeter.

1kW of audio power can deafen you, literally. Be cautious, and stand many many metres away from any audio source of that power before switching it on.